physical chemistry ii

70
1-1 Chapter 1: From Classical to Quantum Mechanics Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions on Concepts Q1.1) How did Planck conclude that the discrepancy between experiments and classical theory for blackbody radiation was at high and not low frequencies? The experimental results and the classical theory agree in the limit of low frequencies and diverge at high frequencies. Q1.2) The inability of classical theory to explain the spectral density distribution of a blackbody was called the ultraviolet catastrophe. Why is this name appropriate? The divergence between the classical theory and the experimental results is very pronounced in the UV region. In particular, because the classical theory predicts that the spectral density increases as 2 ν , the total energy radiated by a blackbody is predicted to be infinite. This prediction was clearly wrong, and the lack of agreement with the experimental results was viewed by scientists of the time as a catastrophe. Q1.3) Why does the analysis of the photoelectric effect based on classical physics predict that the kinetic energy of electrons will increase with increasing light intensity? In the classical theory, more light intensity leads to more energy absorption by the solid and the electrons. At equilibrium, as much energy must leave the surface as is absorbed. Therefore, the electrons will have a greater energy. Classical physicists thought that the greater energy would manifest as a higher speed. In fact, it appears as a larger number of electrons leaving the surface with a frequency determined by the photon energy and the work function of the surface. Q1.4) What did Einstein postulate to explain that the kinetic energy of the emitted electrons in the photoelectric effect depends on the frequency? How does this postulate differ from the predictions of classical physics? Einstein postulated that the energy of light depends on the frequency, whereas in classical theory, the energy depends only on the intensity and is independent of the frequency. He further postulated that the energy of light could only be an integral multiple of . hv Q1.5) Which of the experimental results for the photoelectric effect suggests that light can display particle-like behavior?

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Page 1: Physical Chemistry II

1-1

Chapter 1: From Classical to Quantum Mechanics

Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions on Concepts

Q1.1) How did Planck conclude that the discrepancy between experiments and classical theory for blackbody radiation was at high and not low frequencies? The experimental results and the classical theory agree in the limit of low frequencies and diverge at high frequencies. Q1.2) The inability of classical theory to explain the spectral density distribution of a blackbody was called the ultraviolet catastrophe. Why is this name appropriate? The divergence between the classical theory and the experimental results is very pronounced in the UV region. In particular, because the classical theory predicts that the spectral density increases as 2ν , the total energy radiated by a blackbody is predicted to be infinite. This prediction was clearly wrong, and the lack of agreement with the experimental results was viewed by scientists of the time as a catastrophe. Q1.3) Why does the analysis of the photoelectric effect based on classical physics predict that the kinetic energy of electrons will increase with increasing light intensity? In the classical theory, more light intensity leads to more energy absorption by the solid and the electrons. At equilibrium, as much energy must leave the surface as is absorbed. Therefore, the electrons will have a greater energy. Classical physicists thought that the greater energy would manifest as a higher speed. In fact, it appears as a larger number of electrons leaving the surface with a frequency determined by the photon energy and the work function of the surface. Q1.4) What did Einstein postulate to explain that the kinetic energy of the emitted electrons in the photoelectric effect depends on the frequency? How does this postulate differ from the predictions of classical physics? Einstein postulated that the energy of light depends on the frequency, whereas in classical theory, the energy depends only on the intensity and is independent of the frequency. He further postulated that the energy of light could only be an integral multiple of .hv Q1.5) Which of the experimental results for the photoelectric effect suggests that light can display particle-like behavior?

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Chapter 1/From Classical to Quantum Mechanics

1-2

The fact that at very low intensity the incident light is still able to eject electrons, if its frequency is above the threshold frequency, suggests that the energy of the light can be concentrated in a region of atomic dimensions at the surface of the solid. This suggests particle-like behavior. Q1.6) In the diffraction of electrons by crystals, the volume sampled by the diffracting electrons is on the order of 3 to 10 atomic layers. If He atoms are incident on the surface, only the topmost atomic layer is sampled. Can you explain this difference? Whereas electrons can penetrate through the topmost layer into the solid, He atoms are too large to allow penetration at thermal energies. Therefore electrons sample a number of atomic layers below the surface, and He atoms are only sensitive to the outermost layer. Q1.7) In the double-slit experiment, researchers found that an equal amount of energy passes through each slit. Does this result allow you to distinguish between purely particle-like or purely wave-like behavior? No. This result would be expected for both waves and particles. In the case of particles, the same number would pass through each slit for a large number of incident particles. For a wave, the fraction of the intensity of the wave that passes through each slit would be the same. Q1.8) Is the intensity observed from the diffraction experiment depicted in Figure 1.6 the same for the angles shown in parts (b) and (c)? Yes. For all minima in a graph of intensity versus frequency, the intensity is zero as seen in Figure 1.5. Q1.9) What feature of the distribution depicted as Case 1 in Figure 1.7 tells you that the broad distribution arises from diffraction? The intensity goes through a minimum and increases again on both sides of the maximum. This can only occur through wave interference. Q1.10) Why were investigations at the atomic and subatomic levels required to detect the wave nature of particles? The wavelength of particles with greater mass is so short that particle diffraction could not be observed experimentally. Therefore, the wave-particle duality present even in heavier particles was not observed.

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Chapter 1/From Classical to Quantum Mechanics

1-3

Problems P1.1) The distribution in wavelengths of the light emitted from a radiating blackbody is a sensitive function of the temperature. This dependence is used to measure the temperature of hot objects, without making physical contact with those objects, in a technique called optical pyrometry. In the limit ( ) 1,hc k Tλ >> the maximum in a plot of

( ), versus Tρ λ λ is given by max / 5 .hc kTλ = At what wavelength does the maximum in

( ),Tρ λ occur for T = 450, 1500, and 4500 K?

According to Example Problem 3.1, max 5h ck T

λ = .

34 8 16

max 23 1

6.626 10 J s 2.998 10 m s 6.40 10 m5 1.381 10 J K 450 K

λ− −

−− −

× × ×= = ×× × ×

at 450 K. maxλ for 1500 K and

4500 K is 1.92× 10–6 m and 6.39× 10–7 m, respectively. P1.2) For a monatomic gas, one measure of the “average speed” of the atoms is the root

mean square speed, 1

2 2 3v v ,rmskTm

= = in which m is the molecular mass and k is the

Boltzmann constant. Using this formula, calculate the de Broglie wavelength for He and Ar atoms at 100 and at 500 K.

34

23 1 27 1rms

10

6.626 10 J sv 3 3 1.381 10 J K 100 4.003amu 1.661 10 kg amu

1.26 10 m

h hm k T m K

λ−

− − − −

×= = =× × × × × ×

= ×

for He at 100 K. λ =5.65× 10–11 m for He at 500 K. For Ar, λ = 4.00× 10–11 m and 1.79× 10–11 m at 100 K and 500 K, respectively.

P1.3) Using the root mean square speed, 1

2 2 3v v ,rmskTm

= = calculate the gas

temperatures of He and Ar for which λ = 0.20 nm, a typical value needed to resolve diffraction from the surface of a metal crystal. On the basis of your result, explain why Ar atomic beams are not suitable for atomic diffraction experiments. For He,

( )( )

2

2

234

223 1 27 1 9

3

6.626 10 J s40 K

3 1.381 10 J K 4.003amu 1.661 10 kg amu 0.20 10 m

hTk m λ

− − − − −

=

×= =

× × × × × × ×

For Ar, T = 4.0 K.

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Chapter 1/From Classical to Quantum Mechanics

1-4

The argon temperature is well below its liquifaction temperature at 4 K. It will not be possible to make an atomic beam of Ar atoms with this wavelength with conventional means. P1.4) Electrons have been used to determine molecular structure by diffraction. Calculate the speed of an electron for which the wavelength is equal to a typical bond length, namely, 0.150 nm.

346 1

31 9

6.626 10 J sv 4.85 10 m s9.109 10 kg 0.150 10 m

p hm mλ

−−

− −

×= = = = ×× × ×

P1.5) Calculate the speed that a gas-phase oxygen molecule would have if it had the same energy as an infrared photon (λ = 104 nm), a visible photon (λ = 500 nm), an ultraviolet photon (λ = 100 nm), and an X-ray photon (λ = 0.1 nm). What temperature would the gas have if it had the same energy as each of these photons? Use the root mean

square speed, 1

2 2 3v v ,rmskTm

= = for this calculation.

( )34 8 1

1127 9

2 2 2 6.626 10 Js 2.998 10 m sv 864 m s32.0amu 1.661 10 kg amu 10000 10 m

E h cm m λ

− −−

−− −

× × × ×= = = =× × × ×

for

λ = 104 nm.The results for 500 nm, 100 nm and 0.1 nm are 3.87× 103 m s–1, 8.65× 103 m s–1, and 2.73× 105 m s–1. We calculate the temperature using the formula

( )21 12

1 1

32.0kg mol 864m sv 958K3 3 8.314 J mol K

rmsMTR

− −

− −

×= = =

×

λ = 104 nm. The results for 500 nm, 100 nm and 0.1 nm are 1.92× 104 K, 9.60× 104 K, and 9.56× 107 K. P1.6) Pulsed lasers are powerful sources of nearly monochromatic radiation. Lasers that emit photons in a pulse of 10-ns duration with a total energy in the pulse of 0.10 J at 1000 nm are commercially available. a) What is the average power (energy per unit time) in units of watts (1 W = 1 J/s) associated with such a pulse? b) How many 1000-nm photons are emitted in such a pulse?

a) 7 18

0.10 J 1.0 10 J s1.0 10 s

EPt

−−

∆= = = ×∆ ×

b) 178 1

34 16

0.10 J 5.0 102.998 10 m s6.626 10 J s

1.000 10 m

pulse pulse

photon

E EN cE h

λ

−− −

= = = = ××× ×

×

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Chapter 1/From Classical to Quantum Mechanics

1-5

P1.7) Assume that water absorbs light of wavelength 3.00 × 10–6 m with 100% efficiency. How many photons are required to heat 1.00 g of water by 1.00 K? The heat capacity of water is 75.3 J mol–1 K–1.

,

1 1 6, 19

1 34 8 1

1.00g 75.3J K mol 1.00 K 3.00 10 m 6.31 1018.02g mol 6.626 10 J s 2.998 10 ms

p m

p m

h cE Nh N nC T

C TmNM h c

νλλ − − −

− − −

= = = ∆

∆ × × ×= = = ×× × ×

P1.8) A 1000-W gas discharge lamp emits 3.00 W of ultraviolet radiation in a narrow range centered near 280 nm. How many photons of this wavelength are emitted per second?

1 1 1 118 1

34 8 1

9

3.00 W 1J s W 3.00 W 1J s W 4.23 10 s6.626 10 J s 2.998 10 ms

280 10 m

total

photon

En hcEλ

− − − −−

− −

× ×′ = = = = ×× × ×

×

P1.9) A newly developed substance that emits 225 W of photons with a wavelength of 225 nm is mounted in a small rocket such that all of the radiation is released in the same direction. Because momentum is conserved, the rocket will be accelerated in the opposite direction. If the total mass of the rocket is 5.25 kg, how fast will it be traveling at the end of 365 days in the absence of frictional forces? The number of photons is given by

1 1

20 134 8 1

9

1 J s 1 J s3.00 W 225 WW W 2.547 10 s

6.626 10 J s 2.998 10 ms225 10 m

total

photon

En hcEλ

− −

−− −

× ×′ = = = = ×

× × ××

The momentum of one photon is 34

27 19

6.626 10 J s 2.945 10 kg m s225 10 m

hpλ

−− −

×= = = ××

The force is given by the rate of change of momentum.

( ) 27 1 20 1 7 22.945 10 kg ms 2.547 10 s 7.501 10 kg msd n p

Fdt

− − − − −′= = × × × = ×

The final speed is given by

7 21

07.501 10 kg ms 86400sv v 365days = 4.51ms

5.25kg dayFat tm

− −−×= + = = × ×

P1.10) What speed does a H2 molecule have if it has the same momentum as a photon of wavelength 280 nm?

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Chapter 1/From Classical to Quantum Mechanics

1-6

( )

2 2

2

2

341

127 9

v

6.626 10 J sv 0.707ms2.016amu 1.661 10 kg amu 280 10 m

H H

HH

hp m

hm

λ

λ

−−

−− −

= =

×= = =× × × ×

P1.11) The following data were observed in an experiment of the photoelectric effect from potassium: 10–19 Kinetic Energy (J)

4.49 3.09 1.89 1.34 7.00 3.11

Wavelength (nm)

250 300 350 400 450 500

Graphically evaluate these data to obtain values for the work function and Planck’s constant.

2.5·106 3·106 3.5·106 4·106

1·10-19

2·10-19

3·10-19

4·10-19

E (J)

1 The best-fit line is given by 19 25 13.97362 10 2.11171 10 .E

λ− −= − × + ×

Because the slope is hc, 25

348 1

2.11171 10 J m 7.0 10 J s.2.998 10 m s

h−

−−

×= ≈ ××

The work function is

given by the intercept of the line with the x axis at y = 0. 0

hc Eφλ

= − where

257

0 19

2.1171×10 J m= =5.30×10 m.3.97362×10 J

λ−

−− This gives 194.0 10 J or 2.5eV.φ −≈ ×

P1.12) Show that the energy density radiated by a blackbody

( ) ( )3

3 /0 0

8 1,1

totalhv kT

E T hT d dV c e

π νρ ν ν ν∞ ∞

= =−∫ ∫ depends on the temperature as T 4.

(Hint: Make the substitution of variables / .x h kTν= ) The definite integral 3 4

0

.1 15x

x dxe

π∞

=−∫ Using your result, calculate the energy density radiated by a blackbody

at 800 and 4000 K.

Page 7: Physical Chemistry II

Chapter 1/From Classical to Quantum Mechanics

1-7

3

30

8 1 .1

totalh

kT

E h dV c e

υπ ν ν

=−

∫ Let hx kTν= ; hdx d

kTν=

3 4 4 3 5 4 4

3 3 3 3 30 0

8 1 8 81 151

h xkT

h k T x k Td dxc h c e h ce

υπ ν π πν

∞ ∞

= =−−

∫ ∫

At 800 K, ( ) ( )

5 4 4 5 23 1 4 44 3

3 33 3 34 8 1

8 8 (1.381 10 J K ) (800 K) 3.10 10 J m15 15 6.626×10 J s 2.998 10 m s

totalE k TV h c

π π − −− −

− −

× ×= = = ×× ×

At 4000, ( ) ( )

5 23 1 4 43

3 334 8 1

8 (1.381 10 J K ) (4000 K) 0.194 J m15 6.626 10 J s 2.998 10 m s

totalEV

π −−−

− −

× × ×= =× × × ×

P1.13) The power per unit area emitted by a blackbody is given by

4 8 2 4 with 5.67 × 10 W m K .P Tσ σ − − −= = Calculate the energy radiated by a spherical blackbody of radius 0.500 m at 1000 K per second. What would the radius of a blackbody at 2500 K be if it emitted the same energy as the spherical blackbody of radius 0.500 m at 1000 K?

( ) ( )2 44 8 1 2 4 5 1 =4 0.500m 5.67 10 J s m K 1000 K 1.78 10 J sE A Tσ π − − − − −= × × × = × Because the total energy radiated by the spheres must be equal,

( ) ( )( )

2 4 2 41 1 2 2

2 42 41 1

2 442

4 =4

0.500 m 1000 K= 0.0800 m

2500 K

r T r T

r TrT

π σ π σ

= =

P1.14) In our discussion of blackbody radiation, the average energy of an oscillator

1osc h

kT

hEe

νν=

− was approximated as ( )1 1

oschE kT

hkT

νν

= =+ −

for 1hkT

ν << .

Calculate the relative error = approxE EE

−in making this approximation for

ν = 4 × 1012 s–1 at temperatures of 6000, 2000, and 500 K. Can you predict what the sign of the relative error will be without a detailed calculation?

Page 8: Physical Chemistry II

Chapter 1/From Classical to Quantum Mechanics

1-8

exp 1Relative Error =

exp 1

approx

h kTh

E E kTE

hhkT

νν

νν

− − =

34 12 123 1

34 12 1

23 1

34

6.626 10 J s 4.00 10 s 1.381 10 J K 6000K6.626 10 J s 4.00 10 sexp 1exp 1 1.381 10 J K 6000K

6.626 10 J

exp 1

h kThkT

hhkT

νν

νν

− −− −

− −

− −

× × × − × × − × × × −− × × = ×

12 1

34 12 1

23 1

s 4.00 10 s 16.626 10 J s 4.00 10 sexp

1.381 10 J K 6000K

− −

− −

× × − × × × × ×

Relative Error = 0.0162approxE EE

−= − for T = 6000 K.

The results for 2000 K, and 500 K are –0.0496 and –0.219.

Because Eapprox = kT > E, the relative error = approxE EE

−is always a negative number.

P1.15) The power (energy per unit time) radiated by black body per unit area of surface expressed in units of W m–2 is given by 4P Tσ= with 8 2 45.67 10 W m K .σ − − −= × The radius of the sun is 7.00 × 105 km and the surface temperature is 6000 K. Calculate the total energy radiated per second by the sun. Assume ideal blackbody behavior.

( ) ( )4 2

248 2 4 8

26

4

5.67 10 W m K 6000 K 4 7.00 10 m

4.52 10 W

E PA T rσ π

π− − −

= = ×

= × × × × ×

= ×

P1.16) A more accurate expression for E osc would be obtained by including additional terms in the Taylor-Mclaurin series. The Taylor-Mclaurin series expansion of f(x) in the vicinity of x0 is given by (see Math Supplement)

0 0 0

2 32 3

0 0 0 02 3

( ) 1 ( ) 1 ( )( ) ( ) ( ) ( ) ( ) ... 2! 3! x x x x x x

d f x d f x d f xf x f x x x x x x xd x d x d x= = =

= + − + − + − +

Use this formalism to better approximate E osc by expanding hkTeν

in powers of h kTν out

to ( )3h kTν in the vicinity of 0.h kTν = Calculate the relative error, E kTE

osc

osc

− , if you

Page 9: Physical Chemistry II

Chapter 1/From Classical to Quantum Mechanics

1-9

had not included the additional terms for ν = 1.00 × 1012 s–1 at temperatures of 800, 500, and 250 K.

The Taylor series expansion of hkTeν

is 2 31 11 ...

2 6h h hkT kT kTν ν ν + + + +

. Therefore

including terms up to ( )3hkT

ν ,

2 31 12 6

oschE

h h hkT kT kT

νν ν ν

= + +

2 3 2 31 1 1 12 6 2 6

osc

osc

E kT h hkTE h h h h h h

kT kT kT kT kT kT

ν νν ν ν ν ν ν

− = − + + + +

234 12 1 34 12 1

23 1 23 134 12 1

34 12 1

23 1

6.626 10 J s 1.00 10 s 1 6.626 10 J s 1.00 10 s1.381 10 J K 800 K 2 1.381 10 J K 800 K

6.626 10 J s 1.00 10 s1 6.626 10 J s 1.00 10 s6 1.381 10 J K 800 K

osc

osc

E kTE

− − − −

− − − −− −

− −

− −

× × × × × ×+ × × × ×− = × × × × × ×+ × ×

23 13

234 12 1 34 12 1

23 1 23 134 12 1

3

1.381 10 J K 800K

6.626 10 J s 1.00 10 s 1 6.626 10 J s 1.00 10 s1.381 10 J K 800K 2 1.381 10 J K 800 K

6.626 10 J s 1.00 10 s1 6.626 106

− −

− − − −

− − − −− −

− × × ×

× × × × × ×+ × × × × × × ××+

34 12 1

23 1J s 1.00 10 s

1.381 10 J K 800 K

− −

× × × ×

0.0306osc

osc

E kTE

− = − for 800 K. The corresponding values for 500 K and 250 K are

–0.0495 and –0.102. P1.17) The observed lines in the emission spectrum of atomic hydrogen are given by

( ) ( )1 1 112 2

1

1 1cm cm cm ,HR n nn n

ν − − − = − >

. In the notation favored by spectroscopists,

1 Ehc

νλ

= = and 1109,677cm .HR −= The Lyman, Balmer, and Paschen series refers to

n1 = 1, 2, and 3, respectively, for emission from atomic hydrogen. What is the highest value of and v E in each of these series?

The highest value for ν corresponds to 1n

→ 0. Therefore

1 12

1 cm 109,677 cm1HRν − − = =

or Emax = 2.18× 10–18 J for the Lyman series.

1 12

1 cm 27419 cm2HRν − − = =

or Emax = 5.45× 10–19 J for the Balmer series, and

Page 10: Physical Chemistry II

Chapter 1/From Classical to Quantum Mechanics

1-10

1 12

1 cm 12186cm3HRν − − = =

or Emax = 2.42× 10–19 J for the Paschen series

P1.18) A beam of electrons with a speed of 3.50 × 104 m/s is incident on a slit of width 200 nm. The distance to the detector plane is chosen such that the distance between the central maximum of the diffraction pattern and the first diffraction minimum is 0.500 cm. How far is the detector plane from the slit?

The diffraction minima satisfy the condition sin , 1, 2,...n naλθ = = ± ± and the first

minimum is at sin .aλθ = ± We choose the plus sign (the minus sign gives the distance

from the slit in the opposite direction) giving 34

31 4 1 9

6.626 10 J ssin 0.10392v 9.109 10 kg 3.50 10 ms 200 10 m

5.97 degrees

ha m aλθ

θ

− − −

×= = = =× × × × ×

=

The distance d from the screen and the position of the first minimum s are related by 0.500cm 4.78cm.

tan 0.1045sd

θ= = =

P1.19) If an electron passes through an electrical potential difference of 1 V, it has an energy of 1 electron-volt. What potential difference must it pass through in order to have a wavelength of 0.100 nm?

( )( )

2 22

2

23417

2 1931 10

1 1v2 2 2

6.626 10 J s 1eV2.41 10 J 150.4 eV1.602 10 J2 9.109 10 kg 10 m

e ee e

h hE m mm mλ λ

−−

−− −

= = × =

×= = × × =

×× × ×

The electron must pass through an electrical potential of 150.4 V. P1.20) What is the maximum number of electrons that can be emitted if a potassium surface of work function 2.40 eV absorbs 3.25 × 10–3 J of radiation at a wavelength of 300 nm? What is the kinetic energy and velocity of the electrons emitted?

34 8 1 1919

9

6.626 10 J s 2.998 10 ms 1.602 10 J2.40eV 2.77 10 J300 10 m eV

cE h φλ

− − −−

× × × ×= − = − × = ××

195 1

31

2 2 2.77 10 Jv 7.80 10 m s9.109 10 kg

Em

−−

× ×= = = ××

( )3

1534 8 1 9

3.25 10 J 4.91 106.626 10 J s 2.998 10 m s 300 10 m

total total

photon

E EnE hc λ

− − −

×= = = = ×× × × ×

electrons.

Page 11: Physical Chemistry II

Chapter 1/From Classical to Quantum Mechanics

1-11

P1.21) The work function of platinum is 5.65 eV. What is the minimum frequency of light required to observe the photoelectric effect on Pt? If light with a 150-nm wavelength is absorbed by the surface, what is the velocity of the emitted electrons? a) For electrons to be emitted, the photon energy must be greater than the work function of the surface.

1919

1915 1

34

1.602×10 J5.65 eV 9.05 10 JeV

9.05 10 J 1.37 10 s6.626 10 J s

E h

Eh

ν

ν

−−

−−

= ≥ × = ×

×≥ ≥ ≥ ××

b) The outgoing electron must first surmount the barrier arising from the work function, so not all the photon energy is converted to kinetic energy.

34 8 1

19 199

195 1

31

6.626 10 J s 2.998 10 m s 9.05 10 J 4.19 10 J150 10 m

2 2 4.19 10 Jv = 9.59 10 m s9.11 10 kg

e

e

e

hcE h

Em

ν φ φλ

− −− −

−−

= − = −

× × ×= − × = ××

× ×= = ××

P1.22) X-rays can be generated by accelerating electrons in a vacuum and letting them impact on atoms in a metal surface. If the 1000-eV kinetic energy of the electrons is completely converted to the photon energy, what is the wavelength of the X-rays produced? If the electron current is 1.50 × 10–5 A, how many photons are produced per second?

34 8 1

196.626 10 J s 2.998 10 ms 1.24nm

1.602 10 J1000eVeV

hcE

λ− −

× × ×= = =××

5 113 1

19

current 1.50 10 Cs 9.36 10 s .charge per electron 1.602 10 C

n− −

−−

×= = = ××

P1.23) When a molecule absorbs a photon, both the energy and momentum are conserved. If a H2 molecule at 300 K absorbs an ultraviolet photon of wavelength 100 nm, what is the change in its velocity v∆ ? Given that its average speed is v 3 /rms kT m= , what is v / v ?rms∆ Because momentum is conserved,

3427 1

9

6.626 10 J s 6.626 10 kg m s100 10 mphoton

hpλ

−− −

×= = = ××

Page 12: Physical Chemistry II

Chapter 1/From Classical to Quantum Mechanics

1-12

All this momentum is transferred to the H2 molecule

2

2

27 1

27 11

27

1 13

1 1

3 1

= 6.626 10 kg m s v

6.626 10 kg m sv 1.98 m s2.016 amu 1.661 10 kg/amu

v 1.98 ms 1.98 ms 1.03 10v 3 3 8.314 J mol K 298K

2.016 10 kg mol

H

H

p m

RTM

− −

− −−

− −−

− −

− −

∆ × = ∆

×∆ = =× ×

∆ = = = ×× ×

×

Page 13: Physical Chemistry II

2-1

Chapter 2: The Schrödinger Equation

Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions on Concepts Q2.1) By discussing the diffraction of a beam of particles by a single slit, justify the statement that there is no sharp boundary between particle-like and wave-like behavior. For a low particle energy, corresponding to a long wavelength, the diffraction pattern is clearly resolved. As the energy increases, the wavelength decreases, the diffraction peak moves closer to the central peak and its intensity decreases. The diffraction peak does not disappear with increasing energy; it simply becomes difficult to observe. Q2.2) Why does a quantum mechanical system with discrete energy levels behave as if it has a continuous energy spectrum if the energy difference between energy levels E∆ satisfies the relationship ?E kT∆ << If the difference in energy between levels becomes small compared to kT, the levels become “smeared out” and overlap. When this happens, the levels can no longer be distinguished, and from the viewpoint of the observer, the system has a continuous energy spectrum. Q2.3) Why can we conclude that the wave function ( )( , ) ( ) i E tx t x eψ ψ −= h represents a standing wave? It represents a standing wave because it can be written as the product of a function that depends only on time with a function that depends only on the spatial coordinate. Therefore the nodes do not move with time. Q2.4) Why is it true for any quantum mechanical problem that the set of wave functions is larger than the set of eigenfunctions? The set of all wave functions must satisfy the boundary conditions as well as satisfy the conditions that allow us to interpret the square of the magnitude of the wave function in terms of probability. The set of eigenfunctions must satisfy an additional condition given by the eigenvalue equation. Because some wave functions won’t satisfy the eigenvalue equation, the set of wave functions is larger than the set of eigenfunctions. Q2.5) Is it correct to say that because the de Broglie wavelength of a H2 molecule at 300 K is on the order of atomic dimensions that all properties of H2 are quantized?

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Chapter 2/The Schrödinger Equation

2-2

No. The system must be treated using quantum mechanics if the characteristic size is comparable to the wavelength. For a H2 molecule moving in a one liter container, the wavelength is much smaller than the characteristic dimension of the container. In this case, the translational energy levels are so closely spaced that they can be described classically. Therefore, the pressure exerted by the H2 molecules on the walls of the container can be described classically. Q2.6) In Figure 2.6 the extent to which the approximate and true functions agree was judged visually. How could you quantify the quality of the fit? You could use a function like ( ) ( ) 2

f x g x dx−⎡ ⎤⎣ ⎦∫ , where f(x) and g(x) are the

functions to be compared, as a quantitative measure of how alike the functions are. The value of the integral goes to zero as the two functions become identical. Q2.7) If ( , ) sin( )x t A kx tψ ω= − describes a wave traveling in the plus x direction, how would you describe a wave moving in the minus x direction? Consider the nodes in the function ( , ) sin( ).x t A kx tψ ω= + The wave amplitude

is zero for 2 x t nT

π πλ

⎛ ⎞+ =⎜ ⎟⎝ ⎠

where n is an integer. Solving for x, we obtain the

location of the nodes. 2n tx

Tλ ⎛ ⎞= −⎜ ⎟⎝ ⎠

. We see that x decreases as t increases,

showing that the wave is moving in the direction of negative x. Q2.8) A traveling wave with arbitrary phase φ can be written as

( , ) sin( ).x t A kx tψ ω φ= − + What are the units of φ? Show that φ could be used to represent a shift in the origin of time or distance. The units of φ are radians. We can rewrite this equation either as

( , ) sin( )x t A k x tkφψ ω⎡ ⎤= + −⎢ ⎥⎣ ⎦

or ( , ) sin( )x t A kx t φψ ωω

⎡ ⎤= − −⎢ ⎥⎣ ⎦. The first of

these equations suggests a shift in x, and the second suggests a shift in t. Q2.9) One source emits spherical waves and another emits plane waves. For which source does the intensity measured by a detector of fixed size fall off more rapidly with distance? Why? The intensity of the spherical source falls off more rapidly. The total intensity is the same for all spheres centered at the source. Larger spheres correspond to larger distances, and for a given detector area, the fraction of the intensity in the

area A is 24Arπ

, where r is the radius of the sphere, which is equal to the distance

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Chapter 2/The Schrödinger Equation

2-3

from the source to the detector. Therefore the measured intensity for the spherical source decreases as 1/r2. There is no decrease in intensity for the source emitting plane waves. A well collimated light source, such as a laser, is an approximation to a plane wave source. Q2.10) Distinguish between the following terms applied to a set of functions: orthogonal, normalized, and complete. Two functions φi(x) and φj(x) are orthogonal if ( ) ( )* 0i jx x dxφ φ =∫ , normalized

if ( ) ( ) ( ) ( )* * 1i i j jx x dx x x dxφ φ φ φ= =∫ ∫ , and complete if any well behaved

function ψ(x) can be expanded in terms of the φi(x), ( ) ( )1

m mm

x b xψ φ∞

=

= ∑ .

Problems P2.1) Assume that a system has a very large number of energy levels given by the formula 2 22

0 0 with 2.34 10 J,lε ε ε −= = × where l takes on the integral values 1, 2, 3, … . Assume further that the degeneracy of a level is given by g ll = 2 . Calculate the ratios

5 1 10 1/ and /n n n n for T = 100 K and T = 650 K, respectively.

( ) ( )

( )

( )

20 05 15 5

1 1

225

23 11

225

23 11

10

1

52 5exp exp2 1

2.34×10 J 25 110(100 K) exp 0.0862 1.381 10 J K 100 K

2.34×10 J 25 110(650 K) exp 2.672 1.381 10 J K 650 K

(100 K

n gn g k T k T

nn

nn

nn

ε εε ε

− −

− −

⎡ ⎤− −− −⎡ ⎤ ×⎢ ⎥= =⎢ ⎥ × ⎢ ⎥⎣ ⎦ ⎣ ⎦

⎡ ⎤− × −= =⎢ ⎥× ×⎣ ⎦

⎡ ⎤− × −= =⎢ ⎥× ×⎣ ⎦

( )

( )

227

23 1

2210

23 11

2.34 10 J 100 120) exp 5.2 102 1.381 10 J K 100 K

2.34 10 J 100 120(650 K) exp 0.7572 1.381 10 J K 650 K

nn

−−

− −

− −

⎡ ⎤− × × −= = ×⎢ ⎥× ×⎣ ⎦

⎡ ⎤− × × −= =⎢ ⎥× ×⎣ ⎦

P2.2) Consider a two-level system with 21 21

1 2= 3.10 10 J and 6.10 10 J.ε ε− −× = × If g2

= g1, what value of T is required to obtain 2 1/ n n = 0.150? What value of T is required to obtain 2 1/ n n = 0.999?

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Chapter 2/The Schrödinger Equation

2-4

( )

( )

( )( )

( ) ( )

5 12 2

1 1

5 12 2

1 1

2 2

5 1 1 1

5 1

2 2

1 1

21

2 1 23 1

exp

ln ln

1 ln ln

ln ln

3.00×10 Jfor 0.150 11.381 10 J K ln 1 ln 0.150

n gn g k T

n gn g k T

k g nT g n

Tg nkg n

n n T

ε ε

ε ε

ε ε

ε ε

− −

− −⎡ ⎤= ⎢ ⎥

⎣ ⎦−⎛ ⎞ ⎛ ⎞

= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞= −⎢ ⎥⎜ ⎟ ⎜ ⎟− ⎝ ⎠ ⎝ ⎠⎣ ⎦

−=

⎡ ⎤⎛ ⎞ ⎛ ⎞−⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦

= = =× × −⎡ ⎤⎣ ⎦

15K

( ) ( )21

52 1 23 1

3.00 10 Jfor 0.999 2.17 10 K1.381 10 J K ln 1 ln 0.999

n n T−

− −

×= = = ×

× × ⎡ ⎤⎣ ⎦–

P2.3) To plot Ψ( , ) sin( )x t A kx t= −ω as a function of one of the variables x and t, the

other variable needs to be set at a fixed value, x0 or t0. If ΨΨ( , )

max

x0 0 = –0.280, what is the

constant value of x0 in the upper panel of Figure 2.3? If ΨΨ( , )

max

0 0t = –0.309, what is the

constant value of t0 in the lower panel of Figure 2.3? (Hint: The inverse sine function has two solutions within an interval of 2π. Make sure that you choose the correct one.)

For the traveling wave 3( , ) sin 2 sin 21.46 1.00 10

x t x tx t A AT x

π πλ −

⎛ ⎞⎛ ⎞Ψ = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( )

0 0 0

max

2

( ,0) sin 2 0.284; 21.46 m 1.46 m

0.28 radians or 0.28 = 3.43 radians

6.4 10 m or 0.79 m

x x x

x

π π

π−

⎛ ⎞ ⎛ ⎞Ψ= = −⎜ ⎟ ⎜ ⎟Ψ ⎝ ⎠ ⎝ ⎠

= − − −

= − ×

( )

0 0 03 3

max

5 40

(0, ) sin 2 0.309; 21.00 10 1.00 10

0.31 radians or 0.31 3.45 radians

5.0 10 s or 5.5 10 s

t t tAx s x s

t

π π

π

− −

− −

⎛ ⎞ ⎛ ⎞Ψ − −= = −⎜ ⎟ ⎜ ⎟Ψ ⎝ ⎠ ⎝ ⎠

= − − − =

= × − ×

To decide which of the two values best fits the data in Figure 2.2, it is necessary to plot the data.

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2-5

It is seen that the data are fit by x0 = 0.79 m and t0= –5.5× 10–4 s. Adding any positive or negative integral multiple of λ to x0 or T to t0 will give equally good agreement. P2.4) A wave traveling in the z direction is described by the wave function

1 1 2 2( , ) sin( ) sin( )z t A kz t A kz tω φ ω φΨ = − + + − +x y where x and y are vectors of unit length along the x and y axes, respectively. Because the amplitude is perpendicular to the propagation direction, Ψ( , )z t represents a transverse wave. a) What requirements must A1 and A2 satisfy for a plane polarized wave in the x-z plane? b) What requirements must A1 and A2 satisfy for a plane polarized wave in the y-z plane? c) What requirements must A1 and A2 and φ φ1 2 and satisfy for a plane polarized wave in a plane oriented at 45º to the x-z plane? d) What requirements must A1 and A2 andφ φ1 2 and satisfy for a circularly polarized wave? a) The amplitude along the x axis must oscillate, and the amplitude along the y axis must vanish. Therefore 1 20 and 0.A A≠ = b) The amplitude along the y axis must oscillate, and the amplitude along the x axis must vanish. Therefore 1 20 and 0.A A= ≠

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Chapter 2/The Schrödinger Equation

2-6

c) The amplitude along both the x and y axes must oscillate. Therefore 1 20 and 0A A≠ ≠ . Because they must oscillate in phase, 1 2= .φ φ

d) The amplitude along both the x and y axes must oscillate with the same amplitude. Therefore 1 2 0A A= ≠ . For circularly a polarized wave, the x and y components must be

out of phase by π/2. Therefore 1 2=2πφ φ ± . This can be seen by comparing the x and y

amplitudes for the positive sign.

1 1 1 1

1 1

1 1

1 1

( , ) sin( ) sin( )2

let

( , ) sin( ) sin( )2

sin( ) sin( ) cos cos( )sin2 2

sin( ) c

z t A kz t A kz t

kz kz

z t A kz t A kz t

A kz t A kz t kz t

A kz t A

πω φ ω φ

φπω ω

π πω ω ω

ω

Ψ = − + + − + +

′+ =

′ ′Ψ = − + − +

⎡ ⎤′ ′ ′= − + − + −⎢ ⎥⎣ ⎦′= − +

x y

x y

x y

x y os( )kz tω′−

The x and y amplitudes are π/2 out of phase and the sum of the squares of their amplitudes is a constant as required for a circle.

P2.5) Show that a ibc id

ac bd i bc adc d

++

=+ + −

+b g

2 2

( )

2 2 2 2

ac bd i bc ada ib a ib c id ac bd ibc iadc id c id c id c d c d

+ + −+ + − + + −⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ + − + +⎝ ⎠ ⎝ ⎠

P2.6) Does the superposition ψ ω ω( , ) sin( ) sin( )x t A kx t A kx t= − + +2 generate a standing wave? Answer this question by using trigonometric identities to combine the two terms.

( , ) sin( ) 2 sin( )x t A kx t A kx tψ ω ω= − + + Using the identities

( )( )

sin sin cos cos sin

cos cos cos sin sin

α β α β α β

α β α β α β

± = ±

± = m

the previous equation can be simplified to ( ), sin cos cos sin 2 sin cos 2 cos sin

= 3 sin cos cos sinx t A kx t A kx t A kx t A kx t

A kx t A kx tψ ω ω ω ω

ω ω= − + +

+

Because the wave function cannot be written as a single product of a function that is periodic in length with one that is periodic in time, the nodes will not be stationary. Therefore it is not a standing wave.

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2-7

P2.7) Express the following complex numbers in the form .ire θ

a) 2 4i− b) 6 c) 34+ ii

d) 82 4+−

ii

In the notation ,ire θ 2 2 1 Re= and = cos zr z a bz

θ − ⎛ ⎞= + ⎜ ⎟⎜ ⎟

⎝ ⎠.

a) ( )1 12 4 2 5 exp cos 2 5 exp 0.3525

i i iπ−⎛ ⎞− = =⎜ ⎟⎝ ⎠

b) ( ) ( )16 6exp cos 1 6exp 0i −= =

c) ( )13 1 3 10 1 10exp cos exp 0.3984 4 4 4 410

i i i ii

π−+ ⎛ ⎞= − = =⎜ ⎟⎝ ⎠

d) ( )18 3 17 13 6 13exp cos exp 0.3922 4 5 10 2 25 13

i i i ii

π−+ ⎛ ⎞= + = =⎜ ⎟− ⎝ ⎠

P2.8) Express the following complex numbers in the form a + ib.

a) 22i

b) 2 5 2ei− π

c) eiπ d) 3 25 3

4

+e

To convert to the form a + ib, we use the equations Re cos and Im sinz z z zθ θ= =

a) 22 2 cos 2sin 22 2

ie i iπ π π⎛ ⎞= + =⎜ ⎟

⎝ ⎠

b) 22 5 2 5 cos 2 5 sin 2 52 2

ie i i

π π π− − −⎛ ⎞= + = −⎜ ⎟⎝ ⎠

c) ( )cos sin 1ie iπ π π= + = −

d) 43 2 3 2 3 2 3 2 2 3 2 2cos sin cos

4 4 2 4 25 3 5 3 5 3 5 3 5 3

ie i iπ π π π⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟+ + + + +⎝ ⎠ ⎝ ⎠

( )3 15 3

i= ++

P2.9) Using the exponential representation of the sine and cosine functions

( ) ( )1 1cos and sin2 2

i i i ie e e ei

θ θ θ θθ θ− −= + = − , show that

a) cos sin2 2 1θ θ+ =

b) d

dcos

sinθ

θθ

b g= −

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Chapter 2/The Schrödinger Equation

2-8

c) sin cosθ π θ+FHGIKJ =2

a)

( ) ( )

( ) ( )

2 22 2

2 2 2 2

1 1cos sin2 2

1 12 2 14 4

i i i i

i i i i

e e e ei

e e e e

θ θ θ θ

θ θ θ θ

θ θ − −

− −

⎡ ⎤ ⎡ ⎤+ = + + −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= + + − − + + =

b)

( ) ( )( ) ( )

1cos 1 12 sin

2 2

i i

i i i id e ed

ie ie e ed d

θ θ

θ θ θ θθθ

θ θ

− −

⎡ ⎤+⎢ ⎥⎣ ⎦= = − = − − = −−

c)

( )2 2 2 21 1sin cos2 2 2 2

i i i ii i i iie e e e e e e ei i i

π π π πθ θθ θ θ θπθ θ

⎡ ⎤ ⎡ ⎤+ − + −⎢ ⎥ ⎢ ⎥ − −⎣ ⎦ ⎣ ⎦⎛ ⎞ ⎛ ⎞⎛ ⎞+ = − = − = + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

P2.10) Determine in each of the following cases if the function in the first column is an eigenfunction of the operator in the second column. If so, what is the eigenvalue?

a) sin cosθ φ ∂∂φ

b) 2

2x

e− 1x

dd x

c) sinθ sincos

θθ θ

dd

a) sin cosθ φ φ∂∂

sin cos sin sin .θ φ θ φφ∂

= −∂

Not an eigenfunction

b) 21

2xe− 1 dx d x

2 21 12 21 x xd e e

x d x− −= − Eigenfunction with eigenvalue –1

c) sinθ sincos

dd

θθ θ

sin sin sincos

dd

θ θ θθ θ

= Eigenfunction with eigenvalue +1

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Chapter 2/The Schrödinger Equation

2-9

P2.11) Determine in each of the following cases if the function in the first column is an eigenfunction of the operator in the second column. If so, what is the eigenvalue?

a) x3 3

3

dd x

b) x y xx

yy

∂∂

+∂∂

c) sin cosθ φ ∂∂

2

a) 3x 3

dd x

3

3 6d xd x

= Not an eigenfunction

b) x y x yx y∂ ∂+

∂ ∂

2xy xyx y xyx y

∂ ∂+ =

∂ ∂ Eigenfunction with eigenvalue +2

c) sin cosθ φ 2

2θ∂∂

( )2

2 sin cos sin cosθ φ θ φθ∂

= −∂

Eigenfunction with eigenvalue –1

P2.12) Determine in each of the following cases if the function in the first column is an eigenfunction of the operator in the second column. If so, what is the eigenvalue?

a) 23cos 1θ − 1 sinsin

d dd d

θθ θ θ

⎛ ⎞⎜ ⎟⎝ ⎠

b) 2

2x

e− 2

22

d xdx

c) e i−4 φ dd

2

a) 23cos 1θ − 1 sinsin

d dd d

θθ θ θ

⎛ ⎞⎜ ⎟⎝ ⎠

( ) ( )

( )( )

22

3 2 2 2

2 2

3cos 11 1sin 6 cos sinsin sin

1 6sin 12 cos sin 6sin 12 cossin6 18cos 6 3cos 1

dd dd d d

θθ θ θ

θ θ θ θ θ

θ θ θ θ θθ

θ θ

⎛ ⎞−⎜ ⎟ = −⎜ ⎟⎝ ⎠

= − = −

= − = − −

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Chapter 2/The Schrödinger Equation

2-10

Eigenfunction with eigenvalue –6.

b) 21

2xe− 2

22

d xd x

22 2

12 2 1 12 2 22

xx xd e x e e

d x

−− −− = −

Eigenfunction with eigenvalue –1.

c) 4ie φ− 2

2

ddφ

2 44

2 16i

id e ed

φφ

φ

−−= −

Eigenfunction with eigenvalue –16. P2.13) Determine in each of the following cases if the function in the first column is an eigenfunction of the operator in the second column. If so, what is the eigenvalue?

a) e i x y− +3 2b g ∂∂

2

2x

b) x y2 2+ ( )2 21 x yx x

∂+

c) sin cosθ θ 2sin sin 6sind dd d

θ θ θθ θ⎛ ⎞ +⎜ ⎟⎝ ⎠

a) ( )3 2i x ye− + 2

2x∂∂

( )( )

3 223 2

2 9i x y

i x ye ex

− +− +∂

= −∂

Eigenfunction with eigenvalue –9.

b) 2 2x y+ ( )2 21 x yx x

∂+

( )2 2

2 2 2 21 x yx y x y

x x∂ +

+ = +∂

Eigenfunction with eigenvalue +1.

c) sin cosθ θ 2sin sin 6sind dd d

θ θ θθ θ⎛ ⎞

+⎜ ⎟⎝ ⎠

( )( )

3 2 3

2 3

sin cossin sin 6sin cos sin sin 1 2sin 6sin cos

sin cos 6sin cos 6sin cos sin cos

d d dd d d

θ θθ θ θ θ θ θ θ θ θθ θ θ

θ θ θ θ θ θ θ θ

⎛ ⎞⎡ ⎤+ = − +⎜ ⎟ ⎣ ⎦

⎝ ⎠

= − + =

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Chapter 2/The Schrödinger Equation

2-11

Eigenfunction with eigenvalue +1. P2.14) Which of the following wave functions are eigenfunctions of the operator d/dx? If they are eigenfunctions, what is the eigenvalue? a) a e bex i x− −+3 3 b) sin2 x c) e i x− d) cosa x e) e i x− 2

a) ( )3 3

3 33 3x i x

x ixd a e b e

a e i b ed x

− −− −

+= − − Not an eigenfunction

b) 2sin 2sin cosd x x x

d x= Not an eigenfunction

c) i x

i xd e i ed x

−−= − Eigenfunction with eigenvalue –i

d) cos sind a x a a xd x

= − Not an eigenfunction

e) 2

2

2i x

i xd e i x ed x

−−= − Not an eigenfunction

P2.15) Which of the following wave functions are eigenfunctions of the operator

2 2d dx ? If they are eigenfunctions, what is the eigenvalue? a) 3 3 x i xa e b e− −+ b) sin2 x c) e i x− d) cosa x e) e i x− 2

a) ( )2 3 3

3 32 9 9

x i xx i x

d a e b ea e b e

d x

− −− −

+= − Not an eigenfunction

b) 2 2

2 22

sin 2sin 2 cosd x x xd x

= − + Not an eigenfunction

c) 2

2

i xi xd e e

d x

−−= − Eigenfunction with eigenvalue –1

d) 2

22

cos cosd a x a a xd x

= − Eigenfunction with eigenvalue –a2

e) 2

2 22

22 2 4i x

i x i xd e i e x ed x

−− −= − − Not an eigenfunction

P2.16) If two operators act on a wave function as indicated by $ $AB f xb g , it is important to carry out the operations in succession with the first operation being that nearest to the

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Chapter 2/The Schrödinger Equation

2-12

function. Mathematically, $ $ $ $AB f x A B f xb g b gd i= and $ $ $A f x A A f x2 b g b gd i= . Evaluate the

following successive operations $ $AB f xb g . The operators $A and $B are listed in the first and second columns and f xb g is listed in the third column.

a) dd x

dd x

x ea x2 2

+

b) ∂∂

2

2y ∂

∂ x 2(cos3 )siny

c) ∂∂θ

∂∂

2

2φ cos

sinφθ

a) ( )2

2 2 2

2

2 22 2 2 4 2a x

a x a x a xd x ed d x a x e a x e a e

d x d x d x

⎡ ⎤+⎢ ⎥ ⎡ ⎤= + = + +⎣ ⎦⎢ ⎥⎣ ⎦

b) ( ) [ ]

22 2

2 2

cos 3 sin2 cos 3 sin cos 18cos 3 sin cos

y xy x x y x x

y x y

⎡ ⎤∂∂ ∂= = −⎢ ⎥

∂ ∂ ∂⎢ ⎥⎣ ⎦

c)

2

2 2

coscos cos cossinsin sin

φφ φ θθ

θ φ θ θ θ

⎡ ⎤⎛ ⎞∂ ⎜ ⎟⎢ ⎥∂ ∂ ⎡ ⎤⎝ ⎠⎢ ⎥ = − =⎢ ⎥∂ ∂ ∂ ⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

P2.17) If two operators act on a wave function as indicated by $ $AB f xb g , it is important to carry out the operations in succession with the first operation being that nearest to the function. Mathematically, $ $ $ $AB f x A B f xb g b gd i= and $ $ $A f x A A f x2 b g b gd i= . Evaluate the

following successive operations $ $AB f xb g . The operators $A and $B are listed in the first two columns and f xb g is listed in the third column.

a) dd x

x x e a x− 2

b) x dd x

x e a x− 2

c) yx∂∂

xy∂∂

e a x y− +2 2e j

Note that your answers to parts (a) and (b) are not identical. As we will learn in Chapter 7, the fact that switching the order of the operators x and d dx changes the outcome of the operation $ $AB f xb g is the basis for the Heisenberg uncertainty principle.

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Chapter 2/The Schrödinger Equation

2-13

a) ( )2 2 232 2a x a x a xd x x e x e a x ed x

− − −⎡ ⎤ = −⎣ ⎦

b) ( )2 2 232a x a x a xdx x e x e a x ed x

− − −⎡ ⎤= −⎢ ⎥

⎣ ⎦

c)

( )( ) ( )

( ) ( )

2 2

2 2

2 2 2 22 2 2 2

2

2 4

a x y

a x y

a x y a x y

ey x y a x y e

x y x

a y e a x y e

− +

− +

− + − +

⎡ ⎤∂⎢ ⎥∂ ∂ ⎡ ⎤= − =⎢ ⎥ ⎢ ⎥⎣ ⎦∂ ∂ ∂⎢ ⎥⎣ ⎦

= − +

P2.18) Find the result of operating with 2

22 4d x

dx− on the function

2

.axe− What must the

value of a be to make this function an eigenfunction of the operator?

( )2

2 2 2 2 2 22

2 2 2 2 2 22 4 2 4 4 2 4 1ax

ax ax ax ax ax axd e x e ae x e a x e ae a x edx

−− − − − − −− = − − + = − + −

For the function to be an eigenfunction of the operator, the terms containing

22 axx e− must vanish. This is the case if 1a = ± . P2.19) Find the result of operating with ( )( )( )2 21 2r d dr r d dr r+ on the function

brAe− . What must the values of A and b be to make this function an eigenfunction of the operator?

( )

( )

( )

2 22 2

2 22

2

1 2 1 2

1 22

2 2 2

br br brbr

brbr br

brbr

d dAe Ae d Aer bAr er dr dr r r dr r

AebrAe b r Aer r

Ae b b Aer

− − −−

−− −

−−

+ = − +

= − + +

= − +

To be an eigenfunction of the operator, the terms in brAe

r

must vanish. This requires

that b = 1. There are no restrictions on the value of A.

P2.20) Find the result of operating with 2 2 2

2 2 2

d d ddx dy dz

+ + on the function 2 2 2.x y z+ +

Is this function an eigenfunction of the operator?

( )2 2 2

2 2 22 2 2 6d d d x y z

dx dy dz⎛ ⎞

+ + + + =⎜ ⎟⎝ ⎠

. Therefore, the function is not an eigenfunction of

the operator.

Page 26: Physical Chemistry II

Chapter 2/The Schrödinger Equation

2-14

P2.21) Show that the set of functions ( ) , 0 2 ,i n

n e θφ θ θ π= ≤ ≤ are orthogonal if n and m are integers. To do so, you need to show that the integral

( ) ( )2

*

0

0 for m n d m nπ

φ θ φ θ θ = ≠∫ if n and m are integers.

( ) ( ) ( )

( )( )

( )( )

( ) ( ) ( )

( )

22 2 2*

0 0 0 0

2 0

1

1 1 cos 2 sin 2 1

Because and are integers, is an integer and the argume

i n m i n mim inm n

i n m

d e e d e d ei n m

e e n m n mi n m i n m

n m n m

ππ π πθ θθ θ

π

φ θ φ θ θ θ θ

π π

− −−

⎡ ⎤= = = ⎢ ⎥−⎣ ⎦

⎡ ⎤= − = − + − −⎡ ⎤⎣ ⎦⎣ ⎦− −

∫ ∫ ∫

( ) ( ) ( ) [ ]2

*

0

nts of the sine and cosine functions are integral multiples of 2 .

1 1 0 1 0m n di n m

π

π

φ θ φ θ θ = + − =−∫

P2.22) Show by carrying out the integration that ( ) ( )sin / and cos / ,m x a m x aπ π where m is an integer, are orthogonal over the interval 0 ≤ ≤x a . Would you get the same result if you used the interval 0 3 / 4?x a≤ ≤ Explain your result.

( )2 2

0 0

cos sin sin sin 0 02 2

aa m x m x a m x ad x ma a m a mπ π π π

π π⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎡ ⎤= = − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎣ ⎦⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦∫

3 34 4

0 034

2 2

0

cos sin cos sin

3sin sin 0 02 2 4

a a

a

m x m x m x m xd x d xa a a a

a m x a mm a m

π π π π

π ππ π

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= = − ≠⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

∫ ∫

except for the special case 34m n= where n is an integer. The length of the integration

interval must be n periods (for n an integer) to make the integral zero. P2.23) Normalize the set of functions ( ) , 0 2 .i n

n e θφ θ θ π= ≤ ≤ To do so, you need to multiply the functions by a so-called normalization constant N so that the integral

N N d m nm n* *φ θ φ θ θ

π

b g b g0

2

1z = = for

2 2* * *

0 0

2 1i n i nN N e e d N N d N Nπ π

θ θ θ θ π− = = =∫ ∫ This is satisfied for 12

= and the

normalized functions are ( ) 1 , 0 2 .2

i nn e θφ θ θ π

π= ≤ ≤

Page 27: Physical Chemistry II

Chapter 2/The Schrödinger Equation

2-15

P2.24) In normalizing wave functions, the integration is over all space in which the wave function is defined. The following examples allow you to practice your skills in two- and three-dimensional integration.

a) Normalize the wave function sin sinn x m ya bπ π⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

over the range

0 , 0x a y b≤ ≤ ≤ ≤ . The element of area in two-dimensional Cartesian coordinates is dx dy; n and m are integers and a and b are constants.

b) Normalize the wave function er

a− cos sinθ φ over the interval 0 , 0 , 0 2 .r θ π φ π≤ < ∞ ≤ ≤ ≤ ≤ The volume element in three-dimensional spherical coordinates is r dr d d2 sinθ θ φ . a)

( )

2 2 2

0 0

2 2 2

0 0

2 2 2

0 0

2

1 sin sin

Let , ; then and

1 sin sin

sin sin

Using the standard integral sin

a b

n m

m n

n x m yN d x d ya b

n x m y adu bdwu w dx dya b n m

a bN u w d u d wn m

a bN u d u x w d wn m

ax

π π

π π

π π

π ππ π

π π

π π

⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= = = =

=

⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

∫ ∫

∫ ∫

∫ ∫

∫2

2 2

1 1 sin 22 4

1 1 1 11 sin 2 0 sin 0 sin 2 0 sin 02 4 4 2 4 4

2 2 4

2

dx x axa

a b m nN m nn ma b n m abN N

n m

Nab

π ππ ππ π

π ππ π

= −

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞= − − − × − − −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

= =

=

Page 28: Physical Chemistry II

Chapter 2/The Schrödinger Equation

2-16

b)

( )

2 22 22 2 2 2 2 2 2 2

0 0 0 0 0 0

10

2 3

1 cos sin sin sin cos sin

!Using the standard integral ( 0, positive integer)

2 1 1 1 2!1 sin 2 0 sin 0 cos 0 cos2 4 4 3

r ra a

n axn

N d d e r d r N d d r e d r

nx e dx a na

aN

π π π π

φ θ θ φ θ φ φ θ θ θ

π π π

∞ ∞− −

∞−

+

= =

= >

⎛ ⎞= − − − × − ×⎜ ⎟⎝ ⎠

∫ ∫ ∫ ∫ ∫ ∫

∫3

2 33

3

2 6

6

N a

Na

π

π

⎛ ⎞=⎜ ⎟

⎝ ⎠

=

P2.25) Show that the following pairs of wave functions are orthogonal over the indicated range.

a) ex−

12

2αand ( )

212 22 1 ,

xx e

αα

−− −∞ ≤ < ∞x where α is a constant that is greater than

zero.

b) 0/ 2

0

2 r ar ea

−⎛ ⎞−⎜ ⎟

⎝ ⎠ and r

ae r a

0

2 0− / cosθ over the interval 0 , 0 , 0 2r θ π φ π≤ < ∞ ≤ ≤ ≤ ≤

a)

( ) ( )2 2

2 2 2

2 2

2

1 12 2 22 2

2

0 0

2

0

2 1 2 1 2

4 2 because the integrand is an even function of .

1 3 5 2Using the standard integrals

x x x x x

x x

n ax

x e e d x x e d x x e d x e d x

x e d x e d x x

x e dx

α α α α α

α α

α α α

α

∞ ∞ ∞ ∞− − − − −

−∞ −∞ −∞ −∞

∞ ∞− −

∞−

− = − = −

= −

⋅ ⋅ ⋅ ⋅ ⋅=

∫ ∫ ∫ ∫

∫ ∫

∫( )

( )

2

2 2

1

12

ax

0

1 12 2 2

2

1( 0, positive integer)

2

and e dx4

12 1 = 4 02

n n

x x

na n

a a

a

x e e d xα α

π

π

π πα αα α α

+

∞−

∞− −

−∞

−>

⎛ ⎞= ⎜ ⎟⎝ ⎠

− − =

Page 29: Physical Chemistry II

Chapter 2/The Schrödinger Equation

2-17

b)

0 0

0 0

0 0

2/ 2 / 2 2

0 00 0 0

/ 2 / 2 2

0 00 0

/ 2 / 2 2 2 2

0 00

cos sin 2

2 cos sin 2

1 12 2 cos 0 cos 02 2

r a r a

r a r a

r a r a

r rd d e e r d ra a

r rd e e r d ra a

r re e r d ra a

π π

π

φ θ θ θ

π θ θ θ

π π

∞− −

∞− −

∞− −

⎛ ⎞−⎜ ⎟

⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞

= −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞= − − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

∫ ∫ ∫

∫ ∫

P2.26) Because 0

cos cos 0,d n x m x d x m n

d dπ π⎛ ⎞ ⎛ ⎞ = ≠⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠∫ , the functions

( )cos / for 1, 2, 3, ...n x d nπ = form an orthogonal set. What constant must these functions be multiplied by to form an orthonormal set?

( )

( ) ( )

2 2

00

2

2 2

21 cos cos sin2 4

1 1where we have used the standard integral cos sin 22 4

01 sin 2 sin 0 12 4 2 4 2

2

dd m x m x x d m xN d x N

d d m d

ax dx x axa

d d d dN m Nm m

Nd

π π ππ

ππ π

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

= +

⎡ ⎤= + − − = =⎢ ⎥⎣ ⎦

=

P2.27) Use a Fourier series expansion to express the function ( ) , f x x b x b= − ≤ ≤ in

the form ( ) 01

sin cos .m

n nn

n x n xf x d c db bπ π

=

⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∑ Obtain d0 and the first five

coefficients cn and dn.

Page 30: Physical Chemistry II

Chapter 2/The Schrödinger Equation

2-18

0

2

2 2

1 1( ) 02 2

1 1( ) cos cos

cos sinUsing the standard integral cos

1 cos sin cos

b b

b bb b

nb b

n

d f x dx x dxb b

n x n xd f x dx x dxb b b b

ax x axx axa a

nb n b b n b bd bb n b n b n

π π

ππ ππ π π

− −

− −

= = =

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= +

−⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

∫ ∫

∫ ∫

∫( ) ( ) ( )

( )

2

sin 0

All the = 0 because is an odd function of .

1 1( ) sin sin

cos sinUsing the standard integral sin

n

b b

nb b

n

b n bb bb n b

d f x x

n x n xc f x dx x dxb b b b

x ax axx axa a

c

ππ

π π

− −

⎡ ⎤−⎛ ⎞ ⎛ ⎞⎛ ⎞− − =⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − +

=

∫ ∫

( ) ( ) ( )

( )

2

2 2

2 2

1 sin cos

1 sin cos sin cos

2 sin cos

b

b

n

n

b n x b n xxb n b n b

n b n bb n b b n b b bc b bb n b n b n b n b

b bc n nb n n

π ππ π

π ππ ππ π π π

ππ π

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞−⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤− −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − + −⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞= −⎜ ⎟⎝ ⎠

( )

( ) 1

0 1 5 1 2 3 4 5

2 2cos 1

2 2 2Therefore, 0 and 0. , , , , and 3 2 5

nn

b bc nn n

b b b b bd d d c c c c c

π

ππ π

π π π π π

+

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

⎛ ⎞= = −⎜ ⎟⎝ ⎠

= − = = = − = = − =

P2.28) Carry out the following coordinate transformations. a) Express the point x = 3, y = 2, and z = 1 in spherical coordinates.

b) Express the point r = = =54

34

, ,θ π φ π and in Cartesian coordinates.

a) 2 2 2 2 23 2 1 14r x y z= + + = + + =

1 1

2 2 2

1cos cos 1.30 radians14

zx y z

θ − −= = =+ +

1 1 2tan tan 0.588 radians3

yx

φ − −= = =

b) 3sin cos 5sin cos 2.54 4

x r π πθ φ= = = −

3sin sin 5sin sin 2.54 4

y r π πθ φ= = =

5cos 5cos4 2

z r πθ= = =

Page 31: Physical Chemistry II

Chapter 2/The Schrödinger Equation

2-19

P2.29) Operators can also be expressed as matrices and wave functions as column

vectors. The operator matrix α βδ ε⎛ ⎞⎜ ⎟⎝ ⎠

acts on the wave function ab⎛ ⎞⎜ ⎟⎝ ⎠

according to the

rule a a bb a b

α β α βδ ε δ ε

+⎛ ⎞⎛ ⎞ ⎛ ⎞=⎜ ⎟⎜ ⎟ ⎜ ⎟+⎝ ⎠⎝ ⎠ ⎝ ⎠

. In words, the 2 × 2 matrix operator acting on the two-

element column wave function generates another two-element column wave function. If the wave function generated by the operation is the original wave function multiplied by a constant, the wave function is an eigenfunction of the operator. What is the effect of the

operator 0 11 0⎛ ⎞⎜ ⎟⎝ ⎠

on the column vectors 10⎛ ⎞⎜ ⎟⎝ ⎠

, 01⎛ ⎞⎜ ⎟⎝ ⎠

,11⎛ ⎞⎜ ⎟⎝ ⎠

and 1

1−⎛ ⎞⎜ ⎟⎝ ⎠

? Are these wave

functions eigenfunctions of the operator? See the Math Supplement for a discussion of matrices.

0 1 1 01 0 0 1⎛ ⎞ ⎛ ⎞ ⎛ ⎞

=⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

0 1 0 11 0 1 0⎛ ⎞ ⎛ ⎞ ⎛ ⎞

=⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

0 1 1 11 0 1 1⎛ ⎞ ⎛ ⎞ ⎛ ⎞

=⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

( )0 1 1 1 1

11 0 1 1 1

− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Only 11⎛ ⎞⎜ ⎟⎝ ⎠

and 1

1−⎛ ⎞⎜ ⎟⎝ ⎠

are eigenfunctions with the eigenvalues 1 and –1, respectively.

P2.30) Let 10FHGIKJ and

01FHGIKJ represent the unit vectors along the x and y directions,

respectively. The operator cos sinsin cos

θ θθ θ

−⎛ ⎞⎜ ⎟⎝ ⎠

effects a rotation in the x-y plane. Show that

the length of an arbitrary vector 1 00 1

aa b

b⎛ ⎞ ⎛ ⎞ ⎛ ⎞

= +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

, which is defined as 2 2 ,a b+ is

unchanged by this rotation. See the Math Supplement for a discussion of matrices.

( ) ( )2 2 2 2 2 2

cos sin cos sinsin cos sin cos

The length of the vector is given by

cos sin sin cos ( cos sin 2 sin cos

a a bb a b

a b a b a b ab

θ θ θ θθ θ θ θ

θ θ θ θ θ θ θ θ

− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠ ⎝ ⎠

− + + = + −

( ) ( )

2 2 2 2 1 2

2 2 2 2 2 2 2 2

sin cos 2 sin cos )

cos sin cos sin

a b ab

a b a b

θ θ θ θ

θ θ θ θ

+ + +

= + + + = +

This result shows that the length of the vector is not changed.

Page 32: Physical Chemistry II

4-1

Chapter 4: Using Quantum Mechanics on Simple Systems

Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual.

Questions on Concepts Q4.1) Why are standing-wave solutions for the free particle not compatible with the classical result 0 0vx x t= + ? Because the particle is moving, we must represent it by a traveling wave, for which the nodes move with time. This is not possible using a standing wave, because the nodes do not move with time. Q4.2) Why is it not possible to normalize the free-particle wave functions over the whole range of motion of the particle?

This is the case because ( ) ( )* 2L L ik x ik x

L Lx x dx A A e e dx A A Lψ ψ − +

+ + + +− −= =∫ ∫ diverges as

L→∞. Q4.3) Show that for the particle in the box total energy eigenfunctions,

( ) 2 sinnn xx

a aπψ =

, ( )xψ is a continuous function at the edges of the box. Is ( )d x

d xψ

a

continuous function of x at the edges of the box?

( ) 2 sin 0nn xx

a aπψ = =

at x = 0 and x = a. Because the wave function is zero

everywhere outside the box, it has the value zero at x = 0 and x = a. Therefore, ( )xψ is a continuous function of x because there is no abrupt change in the function at x = 0 or x = a.

( ) 2 2sin cos 1 at 0

cos 1 for zero or even and 1 for odd at .

d x d n x n n x xd x d x a a a a a

n n n x a

ψ π π π

π

= = = =

= − =

Page 33: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-2

( )d xd x

ψ= 0 everywhere outside the box because ψ(x) = 0 outside the box.

Because ( )d xd x

ψcan be one when approaching x = a from inside the box, and is always

equal to zero outside of the box, ( )d xd x

ψis not a continuous function at x = a.

Q4.4) Can the particles in a one-dimensional box, a square two-dimensional box, and a cubic three-dimensional box all have degenerate energy levels? The two and three dimensional boxes can have degenerate energy levels if the lengths along the x and y or x, y, and z directions are the same or if any of the quantities

22 2

2 2 2, , or yx znn na b c

are equal. The one dimensional box levels have only one state per level.

Q4.5) We set the potential energy in the particle in the box equal to zero and justified it by saying that there is no absolute scale for potential energy. Is this also true for kinetic energy?

Yes. 21 v ,2kineticE m= and the velocity is measured relative to the frame of reference. The

same is true if we express the kinetic energy as 2 2

2kinetickEm

= , because the value of

p k= depends on the frame of reference if the observer is moving relative to the source. Q4.6) Why are traveling-wave solutions for the particle in the box not compatible with the boundary conditions? The nodes in traveling waves move with time. This is incompatible with the boundary conditions for the particle in the box. Q4.7) Why is the zero point energy lower for a He atom in a box than for an electron?

The zero point energy 2

28hEma

= varies inversely with the mass, and the mass of a He

atom is much greater than the mass of an electron. Q4.8) Invoke wave-particle duality to address the following question: How does a particle get through a node in a wave function to get to the other side of the box? The question poses a problem only if we try to see the “particle” as a pure particle rather than as a wave-particle. The essence of particle-wave particle duality is that some properties are more easily addressed in a wave picture, and some properties are more

Page 34: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-3

easily addressed in a particle picture. Address this question by looking at the “particle” as a wave. We can easily make a guitar string vibrate over its whole length and still have nodes by holding a finger at the fifth fret. The wave-like properties of the “particle” allow it to have a nonzero probability of being found within any interval dx throughout the box, even if the amplitude of the wave function is zero at a number of special points. Q4.9) What is the difference between probability and probability density? Probability is expressed as ( ) ( )* dψ τ ψ τ τ and is the probability density ( ) ( )*ψ τ ψ τ integrated over the interval .dτ Q4.10) Explain using words, rather than equations, why if ( ) ( ) ( ) ( ), , ,x y zV x y z V x V y V z≠ + +

the total energy eigenfunctions cannot be written in the form ( ) ( ) ( ) ( ), , .x y z X x Y y Z zψ = If the potential has the form ( ) ( ) ( ) ( ), , ,x y zV x y z V x V y V z= + + then the total energy can be written in the form E = Ex + Ey + Ez. This allows the Schrödinger equation to be separated into separate equations for x, y and z and the wave functions to be written in the form ( ) ( ) ( ) ( ), ,x y z X x Y y Z zψ = . However, if ( ) ( ) ( ) ( ), , x y zV x y z V x V y V z≠ + + , this

separation into three separate Schrödinger equations is not possible, and ( ), ,x y zψ cannot be factored into three terms, each of which depends only on one variable. Problems P4.1) Show by examining the position of the nodes that Re A e A ei kx t i kx t

+−

−− −ω ωb g b g and Re

represent plane waves moving in the positive and negative x directions, respectively. The notation Re[ ] refers to the real part of the function in the brackets.

( ) ( )

( )

( )

Re = cos . The nodes of this function occur at

2 12

2 12

i kx tA e kx t

kx t n

x n t

ω ω

πω

π ω

−+

− = +

= + +

Because the position of the nodes is increasing as t increases, this function represents a plane wave moving in the positive x direction.

( ) ( ) ( )

( )

( )

Re cos cos . The nodes of this function occur at

2 12

2 12

i kx tA e kx t kx t

kx t n

x n t

ω ω ω

πω

π ω

− −−

= − − = +

+ = +

= + −

Because the position of the nodes is decreasing as t increases, this function represents a plane wave moving in the negative x direction.

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4-4

P4.2) Show that the energy eigenvalues for the free particle, E km

=2 2

2, are consistent with

the classical result 21 v .2

E m= 2

2

2 2 2

1 v2 2

From the de Brogliie relation,

1 , showing consistency between the classical and quantum result.2 2

pE mm

hp

h kEm m

λ

λ

= =

=

= =

P4.3) Are the total energy eigenfunctions for the free particle in one dimension,

( ) ( )2 22 2

and ,mE mEi x i x

x A e x A eψ ψ+ −

+ −+ −= = eigenfunctions of the one-dimensional linear

momentum operator? If so, what are the eigenvalues?

( )2 22 2

22

2 mE mEi x i x

ikxd mEi A e i A e k A ed x

+ ++

+ + +− = − =

Eigenfunction with eigenvalue k+

( )2 22 2

22

2 mE mEi x i x

ikxd mEi A e i A e k A ed x

− −−

− − −− = − = −

Eigenfunction with eigenvalue k− P4.4) Is the superposition wave function for the free particle

ψ ++

+

−= +x A e A e

i mE x i mE xb g2 2

2 2 an eigenfunction of the momentum operator? Is it an eigenfunction of the total energy operator? Explain your result.

( ) ( )2 2 2 2

2 2

2 2 2 22 2

2 2

2 2

2 2 mE mE mE mEi x i x i x i x

mE mEi x i x

d mE mEi A e A e i A e i A ed x

k A e k A e

+ − + −

+ − + −

+ −

+ −

− + = − +

= −

This function is not an eigenfunction of the momentum operator, because the operation does not return the original function multiplied by a constant.

( ) ( )2 2 2 2

2 2

2 2 2 22 2 2 22 2

2 2 2

2 2

2 2 -2 2 2

mE mE mE mEi x i x i x i x

mE mEi x i x

d mE mEA e A e i A e i A em d x m m

E A e A e

+ − + −

+ − + −

+ −

+ −

− + = − −

= +

This function is an eigenfunction of the total energy operator. Because the energy is

Page 36: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-5

proportional to p2, the difference in sign of the momentum of these two components does not affect the energy. P4.5) Consider a particle in a one-dimensional box defined by

( ) ( )0, 0 and , , 0.V x a x V x x a x= > > = ∞ ≥ ≤ Explain why each of the following unnormalized functions is or is not an acceptable wave function based on criteria such as being consistent with the boundary conditions, and with the association of ψ ψ* x x dxb g b g with probability.

a) A n xa

cos π b) B x x+ 2c h c) C x x a3 −b g d) Dn x

asin π

a) cos n xAaπ is not an acceptable wave function because it does not satisfy the boundary

condition that ( )0 0.ψ = b) ( )2B x x+ is not an acceptable wave function because it does not satisfy the boundary

condition that ( ) 0.aψ = c) ( )3C x x a− is an acceptable wave function. It satisfies both boundary conditions and can be normalized.

d) sin

Dn x

aπ is notan acceptable wave function. It goes to infinity at x = 0 and cannot be

normalized in the desired interval. P4.6) Evaluate the normalization integral for the eigenfunctions of H for the particle in the

box ( ) sinnn xx A

aπψ =

using the trigonometric identity 2 1 cos 2sin

2yy −= .

( )

2 2

0

2 2 2 2

00 0

2 2 2

1 sin

let = ;

1 cos 2 sin 21 sin2 2 4

sin 2 sin 02 2 2

2

a

nn n

n xA dxa

n x ay dx dya n

a a y a y yA y dy A dy An n n

A a A a A an nn n

Aa

ππ π

π

ππ

π π π

π ππ π

=

=

− = = = −

= − − =

=

∫ ∫

Page 37: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-6

P4.7) Use the eigenfunction ( ) +ikx ikxx A e B eψ + −′ ′= rather than ψ x A kx B kxb g = +sin cos to apply the boundary conditions for the particle in the box. a) How do the boundary conditions restrict the acceptable choices for and A B′ ′ and for k? b) Do these two functions give different probability densities if each is normalized?

a)

( ) ( ) ( )( )( )

( )( ) ( )( )

+ cos cos sin sin

cos cos sin sin

0 0, giving . Therefore

cos cos sin sin 2 sin

2 sin 0

ik x ik xx A e B e A k x B k x i A k x B k x

A k x B k x i A k x B k x

A B A B

x A k x A k x i A k x A k x iA k x

a iA k a

ψ

ψψψ

+ −′ ′ ′ ′ ′ ′= = + − + + −

′ ′ ′ ′= + + −

′ ′ ′ ′= + = = −

′ ′ ′ ′ ′= − + + =

′= =

This leads to the same condition on k, namely .k a nπ= The amplitude is now 2 iA′ rather than A. b) The wave function after the imposition of the boundary conditions is

( ) ( ) 2 sin .ik x ik xx A e e iA k xψ + −′ ′= − = After the normalization condition is imposed,

( ) ( ) 22 2iA iAa

′ ′− = . Therefore functions are indistinguishable, because both will give the same

probability densities when normalized. P4.8) Calculate the probability that a particle in a one-dimensional box of length a is found between 0.31a and 0.35a when it is described by the following wave functions:

a) 2 sin xa a

π

b) 2 3sin xa a

π

What would you expect for a classical particle? Compare your results in the two cases with the classical result. a)

( ) ( )

( ) ( )

( ) ( )

2

0.350.352

0.31 0.31

1Using the standard integral sin sin 22 4

2 2sin sin2 4

2 0.35 0.31sin 0.70 sin 0.622 4 2 4

10.04 sin 0.62 sin 0.70 0.0592

aa

a a

yby dy byb

x x a xP dxa a a

a a a aa

π ππ

π ππ π

π ππ

= −

= = −

= − − +

= + − =

Page 38: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-7

b)

( ) ( )

( ) ( )

( ) ( )

2

0.352

0.31

1Using the standard integral sin sin 22 4

2 3 2 0.35 0.31sin sin 2.10 sin 1.862 12 2 12

10.04 sin 1.86 sin 2.10 0.00106

a

a

yby dy byb

x a a a aP dxa a a

π π ππ π

π ππ

= −

= = − − +

= + − =

Because a classical particle is equally likely to be in any given interval, the probability will be 0.04 independent of the energy. In the ground state, the interval chosen is near the maximum of the wave function so that the quantum mechanical probability is greater than the classical probability. For the n = 3 state, the interval chosen is near a node of the wave function so that the quantum mechanical probability is much less than the classical probability. P4.9) Are the eigenfunctions of H for the particle in the one-dimensional box also eigenfunctions of the position operator x? Calculate the average value of x for the case where n = 3. Explain your result by comparing it with what you would expect for a classical particle. Repeat your calculation for n = 5 and, from these two results, suggest an expression valid for all values of n. How does your result compare with the prediction based on classical physics? No, they are not eigenfunctions because multiplying the function by x does not return the function multiplied by a constant. For n = 3,

( ) ( )

( ) ( ) ( )

( ) ( )

* 2

0 0

22

2

2

2

0

2

2

2 3sin

cos 2 sin 2Using the standard integral sin

4 8 4

6 6cos sin2

34 3 48

cos 6 sin 6 c24 72 12

a a

a

xx x x x d x x d xa a

bx x bxxx b x d xb b

x xxx a ax

xa xaa

aaxa

πψ ψ

π π

ππ

π ππ π

= =

= − −

= − −

= − − +

∫ ∫

( ) ( ) 2

2 2 2

os 0 0sin 0 2 1 10 072 12 4 72 72 2

a aaπ π π π

− = − − + − =

For n = 5,

Page 39: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-8

( ) ( )

( ) ( ) ( )

( ) ( )

* 2

0 0

22

2

2

2

0

2

2

2 5sin

cos 2 sin 2Using the standard integral sin

4 8 4

10 10cos sin2

54 5 48

cos 10 sin 1024 200

a a

a

xx x x x d x x d xa a

bx x bxxx b x d xb b

x xxx a ax

xa xaa

aaxa

πψ ψ

π π

ππ

π ππ

= =

= − −

= − −

= − −

∫ ∫

( ) ( ) 2

2 2 2

cos 0 0sin 0 2 1 10 020 200 20 4 200 200

2

aa

aπ π π π π

+ − = − − + −

=

The general expression valid for all states is2ax = . Classical physics gives the same result

because the particle is equally likely to be at any position. The average of all these values is the midpoint of the box. P4.10) Are the eigenfunctions of H for the particle in the one-dimensional box also eigenfunctions of the momentum operator px ? Calculate the average value of px for the case n = 3. Repeat your calculation for n = 5 and, from these two results, suggest an expression valid for all values of n. How does your result compare with the prediction based on classical physics? For n = 3,

( ) ( )

( ) ( ) ( )

( ) ( )

*

0 02

2 2

2 3 3 3sin cos

cosUsing the standard integral sin cos

2cos 3 cos 02 3 2 3 1 1 0

2 2 2 2

a ad i x xp x i x d x d xd x a a a a

bxb x b x d x

bi ip

a a b b a a b b

π π πψ ψ

ππ π

− = − =

=

− − = − = − =

∫ ∫

For n = 5,

( ) ( )

( ) ( ) ( )

( ) ( )

*

0 02

2 2

2 5 5 3sin cos

cosUsing the standard integral sin cos

2cos 5 cos 02 5 2 5 1 1 0

2 2 2 2

a ad i x xp x i x d x d xd x a a a a

bxb x b x d x

bi ip

a a b b a a b b

π π πψ ψ

ππ π

− = − =

=

− − = − = − =

∫ ∫

Page 40: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-9

This is the same result that would be obtained using classical physics. The classical particle is equally likely to be moving in the positive and negative x directions. Therefore the average of a large number of measurements of the momentum is zero for the classical particle moving in a constant potential. P4.11) It is useful to consider the result for the energy eigenvalues for the one-dimensional

box E h nma

nn = =2 2

281 2 3, , , ,...as a function of n, m, and a.

a) By what factor do you need to change the box length to decrease the zero point energy by a factor of 400 for a fixed value of m? b) By what factor would you have to change n for fixed values of a and m to increase the energy by a factor of 400? c) By what factor would you have to increase a at constant n to have the zero point energies of an electron be equal to the zero point energy of a proton in the box?

a) ( )2

2 , 400; 20n

n

aE aE a a′

′ ′= = =

b) ( )

2

2 , 400; 20n

n

E n nE nn′

= = =′′

c)

2 2

27

31

1 1

1.673 10 kg 42.99.109 10 kg

e e p p

pe

p e

m a m a

maa m

=

×= = =×

P4.12) Is the superposition wave function ( ) 2 sin sinn x m xxa a a

π πψ = + an

eigenfunction of the total energy operator for the particle in the box?

( )2 2

2

2 2 2 2

2 2

2 2ˆ sin sin2

2 2sin sin8 8

d n x m xH xm d x a a a a

h n n x h m m xma a a ma a a

π πψ

π π

= − +

= +

Because the result is not the wave function multiplied by a constant, the superposition wave function is not an eigenfunction of the total energy operator.

Page 41: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-10

P4.13) The function ( ) 1 xx Axa

ψ = −

is an acceptable wave function for the particle in the

one dimensional infinite depth box of length a. Calculate the normalization constant A and the expectation values 2 and x x .

2 4 32 2 2

20 0

5 4 3 3 3 3 32 2 2

20 0

3

1 1 2

15 2 3 5 2 3 30

30

a a

a a

x x xA x dx A x dxa a a

x x x a a a aA A Aa a

Aa

= − = − +

= − + = − + =

=

∫ ∫

( ) ( )

( ) ( )

2*

30 0

4 5 6 4 4 4 4

3 2 3 30

22 * 2 2

30 0

5 6 7 5 5 5

3 2 30

30 1

30 2 30 2 304 5 6 6 5 4 60 2

30 1

30 305 3 7 5 3 7

a a

a

a a

a

xx x x x dx x x dxa a

x x x a a a a aa a a a a

xx x x x dx x x dxa a

x x x a a aa a a a

ψ ψ

ψ ψ

= = −

= − + = − + = =

= = −

= − + = − + =

∫ ∫

∫ ∫5 2

3

30 2105 7a a

a=

P4.14) Derive an equation for the probability that a particle characterized by the quantum

number n is in the first quarter ( 04ax≤ ≤ ) of an infinite depth box. Show that this probability

approaches the classical limit as .n → ∞

( ) ( )

( )

2

0.250.252

0 0

1Using the standard integral sin sin 22 4

2 2 2sin sin2 4

2 0 1 1sin sin 0 sin8 4 2 2 4 4 2 2

As , the second term goe

aa

yb y d y b yb

n x x a n xP dxa a a n a

a a n a na n n n

n

π ππ

π ππ π π

= −

= = −

= − − + = −

→ ∞

1s to zero, and the probability approaches . 4

This is the classical value, because the particle is equally likely to be found anywhere in the box.

Page 42: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-11

P4.15) What is the solution of the time-dependent Schrödinger equation ( ),x tΨ for the total

energy eigenfunction ψ π4

2 4xa

xa

b g = FHGIKJsin in the particle in the box model? Write Eω =

explicitly in terms of the parameters of the problem.

( ) ( ) ( )

( ) 2

2 2 2

2 2

4

,

16Because ,8 8

2 4, sin

Etii t

h tima

x t x e x e

n h hEm a m a

xx t ea a

ω

π

ψ ψ ψ

πψ

−−

= =

= =

=

P4.16) For a particle in a two-dimensional box, the total energy eigenfunctions are

( ), sin sinx y

yxn n

n yn xx y Na b

ππψ = .

a) Obtain an expression for ,x yn nE in terms of nx, ny, a, and b by substituting this wave function into the two-dimensional analog of Equation (4.19). b) Contour plots of several eigenfunctions are shown here. The x and y directions of the box lie along the horizontal and vertical directions, respectively. The amplitude has been displayed as a gradation in colors. Regions of positive and negative amplitude are indicated. Identify the values of the quantum numbers nx and ny for plots a–f.

+

+ +

+

+

+

+

+

+

+

+

+

-

-

-

-

-

- -

-

-

-

-

-

a b c

d e f

Page 43: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-12

a) 2 2 2

2 2 2 2

222 2

2 2

4 4sin sin sin sin2

4 4sin sin sin sin2 2

y yx x

y y yx x x

n y n yn x n xm x a a a y a a a

n y n x n yn x n x n xm a a a a m a a a a

π ππ π

π π ππ π π

∂ ∂ − + ∂ ∂

= +

( )2 2 2

2 2

4 sin sin8

x y yxh n n n yn x

m a a a aππ

+ =

Because application of the total energy operator returns the wave function multiplied by a

constant, ( ), sin sinx y

yxn n

n yn xx y Na b

ππψ = is an eigenfunction of the total energy operator.

b) From the result of part (a), ( )2 2 2

, 28x y

x yn n

h n nE

m a+

=

a: nx = 1 ny = 1 b: nx = 2 ny = 3 c: nx = 3 ny = 1 d: nx = 2 ny = 2 e: nx = 1 ny = 5 f: nx = 2 ny = 1 P4.17) Normalize the total energy eigenfunction for the rectangular two-dimensional box,

( ), , sin sinx y

yxn n

n yn xx y Na b

ππψ =

in the interval 0 , 0 .x a y b≤ ≤ ≤ ≤

( ) ( )

( )

( )

* 2 2 2

0 0 0 0

2

2 2 2

0 0

2

1 , , sin sin

sin 2Using the standard integral sin

2 4

sin sin

sin sin 0 sin s2 4 2 4

a b a byx

a byx

x yx y

n yn xx y x y d x d y N d x d ya b

xxx d x

n yn xN d x d ya b

a a b aN n nn n

ππψ ψ

αα

αππ

π ππ π

= =

= −

= − − × − −

∫ ∫ ∫ ∫

∫ ∫

( )

( )

2in 04

4 4 and , sin sin yx

abN

n yn xN x yab ab a b

ππψ

=

= =

P4.18) Consider the contour plots of Problem P4.16. a) What are the most likely area or areas dxdy to find the particle for each of the eigenfunctions of H depicted in plots a–f?

Page 44: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-13

b) For the one-dimensional box, the nodes are points. What form do the nodes take for the two-dimensional box? Where are the nodes located in plots a–f? How many nodes are there in each contour plot? a) What is the most likely point to find the particle for each of the eigenfunctions of H depicted in plots a–e? The most likely location of the particle is the set of points at which the wave function has its largest positive or negative value as indicated by the gray scale. There are 1, 6, 3, 4, 5, and 2 equivalent points in plots a–e respectively. b) For the one-dimensional box, the nodes are points. What form do the nodes take for the two-dimensional box? Where are the nodes located in plots a–e? How many nodes are there in each contour plot? The nodes are lines that lie between values at which the absolute magnitude of the wave function reaches its maximum value. There are 0, 3, 2, 2, 4, and 1 nodes in plots a–e, respectively. P4.19) Show by substitution into Equation (4.19) that the eigenfunctions of H for a box with lengths along the x, y, and z directions of a, b, and c, respectively, are

( ), , , , sin sin sinx y z

yx zn n n

n yn x n zx y z Na b c

ππ πψ =

b) Obtain an expression for , ,x y zn n nE in terms of nx, ny, nz and a, b, and c. a)

2 2 2 2

2 2

2 2

2

sin sin sin sin sin sin2 2

sin sin sin2

y yx xz z

yx z

n y n yn x n xn z n zN Nm x a b c m y a b c

n yn x n zNm z a b c

π ππ ππ π

ππ π

∂ ∂ − − ∂ ∂ ∂ − ∂

=222 2

22

sin sin sin sin sin sin2 2

sin sin sin2

y y yx x xz z

yxz z

n y n x n yn x n x n xn z n zN Nm a a b c m b a b c

n yn xn x n zNm c a b c

π π ππ π ππ π

πππ π

+

+

22 22

2 2 2 sin sin sin8

y yx xz zn n yn n xn n zh N

m a b c a b cππ π

= + +

Page 45: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-14

Because application of the total energy operator returns the wave function multiplied by a

constant, ( ), , , , sin sin sinx y z

yx zn n n

n yn x n zx y z Na b c

ππ πψ = is an eigenfunction of the

total energy operator.

b) Based on the result from part (a), 22 2 2

, , 2 2 28x y z

yx zn n n

nh n nEm a b c

= + +

P4.20) Normalize the total energy eigenfunctions for the three-dimensional box in the interval 0 , 0 , 0 .x a y b z c≤ ≤ ≤ ≤ ≤ ≤

( ) ( )

( )

*

0 0 0

2 2 2 2

0 0 0

2

2 2 2 2

0 0

1 , , , ,

sin sin sin

sin 2Using the standard integral sin

2 4

1 sin sin sin

a b c

a b cyx z

a byx z

x y z x y z d x d y d z

n yn x n zN d x d y d za b b

xxx d x

n yn x n zN d x d ya b b

ψ ψ

ππ π

αα

αππ π

=

=

= −

=

∫ ∫ ∫

∫ ∫ ∫

∫ ∫

( ) ( ) ( )

( )

0

2 21 sin sin 0 sin sin 0 sin sin 02 4 2 4 2 4 8

8 8 and , sin sin sin

c

x y zx y

yx z

d z

a a b a c a abcN n n n Nn n r

n yn x n zN x yabc abc a b c

π π ππ π π

ππ πψ

= − − × − − × − − = = =

P4.21) In discussing the Boltzmann distribution in Chapter 2, we used the symbols gi and gj to indicate the degeneracies of the energy levels i and j. By degeneracy, we mean the number of distinct quantum states (different quantum numbers) all of which have the same energy.

a) Using your answer to Problem P4.16a, what is the degeneracy of the energy level 58

2

2

hma

for

the square two-dimensional box of edge length a?

b) Using your answer to Problem P4.19b, what is the degeneracy of the energy level 98

2

2

hma

for

a three-dimensional cubic box of edge length a?

a) What is the degeneracy of the energy level 2

2

58

hm a

for the square two-dimensional box?

The only pairs nx, ny that satisfy the equation 2 2 5x yn n+ = are 2, 1 and 1, 2. Therefore the degeneracy of this energy level is 2.

b) What is the degeneracy of the energy level 2

2

98

hm a

for a three-dimensional cubic box of edge

length a?

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4-15

The only trios of nonzero numbers n, q, r that satisfy the equation 2 2 2 9x y zn n n+ + = are 2, 2, 1; 2, 1, 2; and 1, 2, 2. Therefore the degeneracy of this energy level is 3. P4.22) This problem explores under what conditions the classical limit is reached for a macroscopic cubic box of edge length a. An argon atom of average translational energy 3/2 kT is confined in a cubic box of volume V = 0.500 m3 at 298 K. Use the result from Equation (4.25) for the dependence of the energy levels on a and on the quantum numbers nx, ny, and nz. a) What is the value of the “reduced quantum number” 2 2 2

x y zn n nα = + + for T = 298 K?

b) What is the energy separation between the levels a and a + 1? (Hint: Subtract Ea+1 from Ea before plugging in numbers.

c) Calculate the ratio E E

k Tα α+ −1 and use your result to conclude whether a classical or

quantum mechanical description is appropriate for the particle.

( )

( )

( )

( )( )

22 2 2

, , 2

2, ,2 2 2

2

2, ,2 2 2

2

227 1 3 23 13

234

10

88

8

8×39.95amu 1.661x10 kg(amu) 0.500 m 1.5 1.381 10 J K 298 K

6.626 10 J s

6.86 10

x y z

x y z

x y z

n n n x y z

n n nx y z

n n nx y z

hE n n nm a

m a En n n

hm a E

n n nh

α

α

α

− − − −

= + +

+ + =

= + + =

× × × × × ×=

×

= ×

b) What is the energy separation between the levels α and α + 1? (Hint: Subtract Eα+1 from Eα before plugging in numbers.

[ ]( ) ( )

( ) ( )( )

222 2

1 2 2

234 1031

227 1 3 3

2 11

8 8

6.626 10 Js 2 6.86 10 +1= 1.80 10 J

8 39.95amu 1.661 10 kg(amu) 0.500 m

hhE Em a m aα α

αα α+

−−

− −

+− = + − =

× × ×= ×

× × × ×

c) Calculate the ratio 1E Ek T

α α+ − and use your result to conclude whether a classical or quantum

mechanical description is appropriate for the particle.

3171

23 1

1.80 10 J 4.44 101.361 10 J K 298 K

E Ek T

α α−

−+− −

− ×= = ×× ×

Page 47: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-16

Because ∆E<<kT, a classical description is appropriate. P4.23) Generally, the quantization of translational motion is not significant for atoms because of their mass. However, this conclusion depends on the dimensions of the space to which they are confined. Zeolites are structures with small pores that we describe by a cube with edge length 1 nm. Calculate the energy of a H2 molecule with nx = ny = nz = 10. Compare this energy to kT at T = 300 K. Is a classical or a quantum description appropriate?

( )

( ) ( )( )

22 2 2

, , 2

234 2 2 221

227 1 9

8

6.626 10 J s 10 10 104.92 10 J

8×2.016amu 1.661x10 kg(amu) 10 m

x y zn n n x y zhE n n nm a

−−

− − −

= + +

× + += = ×

× ×

Using the results of P4.22, the ratio of the energy spacing between levels and kT determines if a classical or quantum description is appropriate.

[ ]( ) ( ) ( )

( ) ( )( )

222 2 2 2 2

1 2 2

23422

227 1 9

221

23 1

2 11 where

8 8

6.626 10 Js 300+1 = 3.00 10 J

8 2.016amu 1.661 10 kg(amu) 10 m

3.00 10 J 0.0731.361 10 J K 298 K

x y z

hhE E n n nm a m a

E EkT

α α

α α

αα α α+

− − −

−+

− −

+− = + − = = + +

×= ×

× × × ×

− ×= =× ×

Because this ratio is not much smaller than one, a quantum description is appropriate. P4.24) Two wave functions are distinguishable if they lead to a different probability density. Which of the following wave functions are distinguishable from sin cos ?iθ θ+

a) ( ) 2 2sin cos2 2

i iθ θ

+ +

b) ( )sin cosi iθ θ+

c) ( )sin cosiθ θ− + d) ( ) 2 2sin cos2 2

i iθ θ

+ − +

e) ( )sin cosiθ θ− f) ie θ

Two wave functions 1 2 and ψ ψ are indistinguishable if 1 2

* *1 2.ψ ψ ψ ψ= For the wave function in

the problem, ( )( ) ( )2 2sin cos sin cos sin cos 1i iθ θ θ θ θ θ− + = + = .

a) ( ) ( ) ( )2 22 2 2 2sin cos sin cos sin cos 12 2 2 2

i i i iθ θ θ θ θ θ

− − + + = + =

indistinguishable.

Page 48: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-17

b) ( ) ( ) ( )2 2sin cos sin cos sin cos 1i i i iθ θ θ θ θ θ− − + = + = indistinguishable c) ( ) ( ) ( )2 2sin cos sin cos sin cos 1i iθ θ θ θ θ θ− − − + = + = indistinguishable

d) ( ) ( ) ( )2 22 2 2 2sin cos sin cos sin cos 12 2 2 2

i i i iθ θ θ θ θ θ

− − − + − + = + =

indistinguishable e) ( ) ( ) ( )2 2sin cos sin cos sin cos 1i iθ θ θ θ θ θ+ − = + = indistinguishable f) 1i ie eθ θ− = indistinguishable P4.25) Suppose that the wave function for a system can be written as

ψ φ φ φx x x i xb g b g b g b g= + + +12

14

3 241 2 3 and that φ φ φ1 2 3x x xb g b g b g, and are normalized

eigenfunctions of the operator Ekinetic with eigenvalues E1, 3E1, and 7E1, respectively. a) Verify that ( )xψ is normalized. b) What are the possible values that you could obtain in measuring the kinetic energy on identically prepared systems? c) What is the probability of measuring each of these eigenvalues? d) What is the average value of Ekinetic that you would obtain from a large number of measurements? a) We first determine if the wave function is normalized.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

1 2 3

1 1 3

2 3

* * * *1 2 3

* * *2 3 1 1

* *3 2

1 1 3 2 3 24 16 4 4

1 3 2 3 2 1 4 16 16 23 2 3 2

8 8

i ix x dx x x dx x x dx x x dx

i ix x dx x x dx x x dx x

i ix x dx x

ψ ψ φ φ φ φ φ φ

φ φ φ φ φ φ φ

φ φ φ φ

− += + +

+ −+ + +

+ −+ +

∫ ∫ ∫ ∫

∫ ∫ ∫

∫ ( )x dx∫ All but the first three integrals are zero because the functions ( ) ( ) ( )1 2 3, , and x x xφ φ φ are orthogonal. The first three integrals have the value one, because the functions are normalized. Therefore,

( ) ( )* 1 1 3 2 3 2 1 1 11 14 16 4 4 4 16 16

i ix x dxψ ψ − += + + = + + =

Page 49: Physical Chemistry II

Chapter 4/Using Quantum Mechanics on Simple Systems

4-18

b) The only possible values of the observable kinetic energy that you will measure are those corresponding to the finite number of terms in the superposition wave function. In this case, the only values that you will measure are E1, 3E1, and 7E1. c) For a normalized superposition wave function, the probability of observing a particular eigenvalue is equal to the square of the magnitude of the coefficient of that kinetic energy eigenfunction in the superposition wave function. These coefficients have been calculated above. The probabilities of observing E1, 3E1, and 7E1 are ¼, 1/16, and 11/16, respectively. d) The average value of the kinetic energy is given by

1 1 1 11 1 113 7 5.254 16 16i iE PE E E E E= = + + =∑

Page 50: Physical Chemistry II

7-1

Chapter 7: A Quantum Mechanical Model for the Vibration and Rotation of Molecules

Problem numbers in italics indicate that the solution is included in the Student’s Solutions Manual. Questions on Concepts Q7.1) What is the functional dependence of the total energy of the quantum harmonic oscillator on the position variable x? It is independent of x. Both the kinetic and potential energy depend on x, but the total energy is independent of x. Q7.2) The two linearly independent total energy eigenfunctions for rotation in two

dimensions are Φ Φ+ −−= =φ

πφ

πφ φb g b g1

212

e eim iml l and . What is the difference in

motion for these two solutions? Explain your answer. From Section 7.4, the angular momentum operator for rotation in two dimensions

is l iφ∂

= −∂

h if we apply this operator to the two eigenfunctions

Φ Φ+ −−= =φ

πφ

πφ φb g b g1

212

e eim iml l and ,

( ) ( ) ( )

( ) ( ) ( )

ˆ = and 2

ˆ = 2

l

l

imll

imll

ml i e m

ml i e m

φ

φ

φ φ φφ π

φ φ φφ π

+ + +

−− − −

∂Φ = − Φ = Φ

∂−∂

Φ = − Φ = − Φ∂

hh h

hh h

Therefore, the direction of rotation is opposite for these two eigenfunctions. Q7.3) Spatial quantization was discussed in Supplemental Section 7.8. Suppose that we have a gas consisting of atoms, each of which has a nonzero angular momentum. Are all of their angular momentum vectors aligned? No. Only the application of an eternal field that makes the energy depend on the component of the angular momentum will lead to spatial quantization. Even in this case, not all molecules will have the same component of angular momentum along the field direction. The number of molecules with a particular value for the component of angular momentum along the field direction will be determined by the Boltzmann distribution.

Page 51: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-2

Q7.4) Why can the angular momentum vector lie on the z axis for two-dimensional rotation in the x-y plane but not for rotation in three-dimensional space? For rotation in two dimensions, the angular momentum vector has only one component, and the vector is perpendicular to the plane of rotation. For rotation in three dimensions, the angular momentum vector has three components. Because the corresponding angular momentum operators do not commute, all three components of the angular momentum can’t be known simultaneously. Therefore, the angular momentum vector can’t lie on the z axis, because this requires that the x and y components are known to be zero, and the z component is known through the value of L2. Q7.5) The zero point energy of the particle in the box goes to zero as the length of the box approaches infinity. What is the appropriate analogue for the quantum harmonic oscillator? The force constant k goes to zero. This corresponds in a diatomic molecule to a very weak bond, such as for Ar2. Q7.6) Explain in words why the amplitude of the total energy eigenfunctions for the quantum mechanical harmonic oscillator increases with |x| as shown in Figure 7.4. The velocity of the particle goes through zero at the classical turning point. Therefore, it spends more time near the turning point than in the same interval near x = 0. Therefore, the probability density must be highest at the classical turning point. Q7.7) Why does the energy of a rotating molecule depend on l, but not on ml? The energy depends only on the frequency of rotation, and not on the orientation of the rotation axis. The quantum number l determines the speed, and ml determines the orientation of the rotational axis. Q7.8) Are the real functions listed in Equations (7.34) and (7.35) eigenfunctions of lz? Justify your answer. No. They are formed by the combination of and .−l lm m

l lY Y Therefore, the

functions are eigenfunctions of 2ˆˆ and ,H l but not of ˆ .zl Q7.9) Why is only one quantum number needed to characterize the eigenfunctions for rotation in two dimensions, whereas two quantum numbers are needed to characterize the eigenfunctions for rotation in three dimensions? For rotation in two dimensions, the angular momentum vector is perpendicular to the plane of rotation, and therefore has only one component. Only one quantum

Page 52: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-3

number is needed to describe this vector. By contrast, the angular momentum vector for 3-D rotation has three components. However, the commutation relations say that we can only know the magnitude of the vector and one of its components simultaneously. Two quantum numbers are needed to describe this vector. Q7.10) What makes the z direction special so that we say that 2l , $H , and zl commute, whereas the individual angular momentum operators zl , yl and xl don’t commute? There is nothing special about the z direction. We choose the z direction because the angular momentum operator for this component has a simple form. We can rotate our coordinate system to make any direction that we choose to lie along the z axis. The important point is that we can know the magnitude of the angular momentum and only one of its components. Problems P7.1) The force constant for a H19F molecule is 966 N m–1. a) Calculate the zero point vibrational energy for this molecule for a harmonic potential. b) Calculate the light frequency needed to excite this molecule from the ground state to the first excited state. a)

134

127 1

191

200 1

1 3 966 N m1+ = 1.055 10 J s 1.0078 18.99842 2 1.66 x10 kg amu1.0078+18.9984

1.23 10 J

1 1 4.10 10 J2 3

−−

− −

⎛ ⎞= × × ×⎜ ⎟ ×⎝ ⎠ ×

= ×

⎛ ⎞= = = ×⎜ ⎟⎝ ⎠

h

kE h

E

kE E

µ

µ

b) 19 20

14 11 034

1.23 10 J 4.10 10 J = 1.24 10 s6.626 10 J s

− −−

− × − ×= = ×

×E E

P7.2) By substituting in the Schrödinger equation for the harmonic oscillator, show that the ground-state vibrational wave function is an eigenfunction of the total energy operator. Determine the energy eigenvalue.

Page 53: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-4

( ) ( ) ( )

22

2 2

2 2

22 2

2

1 4 11 4 122 2

1 4 1 41 12 2 2 22 2

2

1 4 1 41 12 22 22 2

2 2

2 2 2 2

2 2

nn n n

xx

x x

x x

d x k x x E xd x

d x ed ek x k xe e

d x d x

k xe x e

αα

α α

α α

ψψ ψ

µ

αα απα απ

µ π µ π

α α αα αµ π π π

−−

− −

− −

− + =

⎧ ⎫⎪ ⎪⎛ ⎞⎛ ⎞ ⎨ ⎬⎜ ⎟⎜ ⎟ ⎝ ⎠⎪ ⎪⎛ ⎞ ⎛ ⎞⎩ ⎭⎝ ⎠− + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎧ ⎫⎪ ⎪⎛ ⎞ ⎛ ⎞ ⎛= − +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠ ⎝⎪ ⎪⎩ ⎭

h

h h

h

( )

2

2 2 2 2

2 2

1 4 12

1 4 1 4 1 4 1 41 1 1 12 2 2 2 22 22 2 2 2

1 4 1 41 122 2

1 1 12

2 2 2

with 2 2 2

x

x x x x

x x

e

xe x e e e

k k ke e E x E

α

α α α α

α α

α α α α αα α αµ π π µ π µ π

µ α α ψµ π µ π µ

− − − −

− −

⎞⎟⎠

⎧ ⎫⎪ ⎪⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + =⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎪ ⎪⎩ ⎭

⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

h h h

h h h

h

P7.3) Show by carrying out the appropriate integration that the total energy

eigenfunctions for the harmonic oscillator ( )21 4 1

20

xx e

ααψπ

−⎛ ⎞= ⎜ ⎟⎝ ⎠

and

ψ απ

αα

2

1 42

12

42 1

2

x x exb g c h= FHG

IKJ −

−are orthogonal over the interval −∞ < < ∞x and that

ψ 2 xb g is normalized over the same interval. In evaluating integrals of this type,

f x d xb g−∞

∞z = 0 if f(x) is an odd function of x and f x d x f x d xb g b g−∞

∞ ∞z z= 20

if f(x) is an

even function of x. 2

2

210

12

0

1 3 5 (2 1)We use the standard integrals and2

4

n a xn n

ax

nx e d xa a

e dxa

π

π

∞ • • • • •−+

∞−

−=

⎛ ⎞= ⎜ ⎟⎝ ⎠

( ) ( ) ( )

( ) ( )

2 2

2 2

1 4 1 41 1* 2 2 22 0

1 4 1 42 22 2

2 2 0

1 42

2

2 14

2 1 2 2 14 4

1 12 04 4 2

x x

x x

x x d x x e e d x

x e d x x e d x

α α

α α

α αψ ψ απ π

α αα απ π

α π παπ α α α

∞ ∞ − −

−∞ −∞

∞ ∞− −

−∞

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞= − =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

∫ ∫

∫ ∫

Page 54: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-5

( ) ( ) ( ) ( )

( )

2 2

2

1 4 1 41 1* 2 22 22 2

1 22 4 2

0

1 2 1 22

3 2 2

2 1 2 14 4

2 4 4 14

3 1 1 3 12 4 4 2 1 14 2 2 2 4 2 2

x x

x

x x d x x e x e d x

x x e d x

α α

α

α αψ ψ α απ π

α α απ

α π π π α πα απ α α α α α π α

∞ ∞ − −

−∞ −∞

∞ −

⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= − +⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − + = − + =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

∫ ∫

P7.4) Evaluate the average kinetic and potential energies, E Ekinetic potential and , for the ground state (n = 0) of the harmonic oscillator by carrying out the appropriate integrations.

2

2

210

12

0

1 3 5 (2 1)We use the standard integrals and2

4

n a xn n

ax

nx e d xa a

e dxa

π

π

∞ • • • • •−+

∞−

−=

⎛ ⎞= ⎜ ⎟⎝ ⎠

( ) ( )

2 2

* 20 0

1 2 1 22 2

01 2

12

1 2

1 1 =4 4 4

potential

x x

E x k x x dx

k x e dx k x e dx

kk k

α α

ψ ψ

α απ π

α ππ α α α µ

∞ ∞− −

−∞

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ = =⎜ ⎟⎝ ⎠

∫ ∫h

( ) ( )

( )

2 2

2

2 2*0 02

1 4 1 41 12 22 2

2

1 222

0

2

2

kinetic

x x

x

E x x dxx

e e dxx

e x dx

α α

α

ψ ψµ

α απ µ π

α α αµ π

∞− −

−∞

∞−

⎛ ⎞∂= −⎜ ⎟∂⎝ ⎠

⎛ ⎞∂⎛ ⎞ ⎛ ⎞= −⎜ ⎟⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞= − −⎜ ⎟⎝ ⎠

h

h

h

1 22 2

2

2

4 2 4

4 4

k k

α α π α π αµ π α α µ

µµ µ

⎛ ⎞⎛ ⎞= − − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= =

h h

h h

h

Page 55: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-6

P7.5) Evaluate the average kinetic and potential energies, E Ekinetic potential and , for the second excited state (n = 2) of the harmonic oscillator by carrying out the appropriate integrations.

2

2

210

12

0

1 3 5 (2 1)We use the standard integrals and2

4

n a xn n

ax

nx e d xa a

e dxa

π

π

∞ • • • • •−+

∞−

−=

⎛ ⎞= ⎜ ⎟⎝ ⎠

( ) ( ) ( )

( ) ( )

( )

2 2

2

*2 2

1 4 1 41 12 2 22 2

1 222 2

1 2 1 2 14 2 4

1 2 12 4

potential

x x

x

E x V x x dx

x e k x x e dx

k x x e dx

α α

α

ψ ψ

α αα απ π

α απ

∞− −

−∞

∞−

−∞

=

⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= −⎜ ⎟⎝ ⎠

( ) 21 2

2 6 4 2

0

1 22

4 3 3 2 2

4 44

15 3 1 4 44 2 2 2

5 5 =4 4

xk x x x e dx

k

k k

αα α απ

α π π πα απ α α α α α α

α µ

∞−⎛ ⎞= − +⎜ ⎟

⎝ ⎠

⎛ ⎞⎛ ⎞= − +⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

h

( ) ( )

( ) ( )

( ) ( )

2 2

2 2

2 2*2 22

1 4 1 41 12 22 22 2

2

1 4 1 41 122 2 4 22 2

2

2 1 2 14 2 4

2 1 2 11 54 2 4

kinetic

x x

x x

E x x dxx

x e x e dxx

x e x x e dx

α α

α α

ψ ψµ

α αα απ µ π

α αα α α απ µ π

∞− −

−∞

∞− −

−∞

⎛ ⎞∂= −⎜ ⎟∂⎝ ⎠

⎛ ⎞∂⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞= − − − +⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

h

h

h

( )( )

( )

2

2

1 222 2 4 2

0

1 223 6 2 4 2

0

2 1 2 11 54

4 24 21 54

x

x

x x x e dx

x x x e dx

α

α

αα α α αµ π

αα α α αµ π

∞−

∞−

⎛ ⎞⎛ ⎞= − − − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= − − + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

h

h

1 223 2

4 3 3 2 2

2

15 3 21 14 24 54 2 2 2 2

5 54 4

k

α π π π πα α α αµ π α α α α α α α

αµ µ

⎛ ⎞⎛ ⎞⎛ ⎞= − − + −⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

= =

h

hh

Page 56: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-7

P7.6) Evaluate the average vibrational amplitude of the quantum harmonic oscillator about its equilibrium value, x , for the ground state (n = 0) and first two excited states (n = 1 and n = 2). Use the hint about evaluating integrals in Problem P7.3.

( ) ( )2*n n nx x x d x x d xψ ψ ψ

∞ ∞

−∞ −∞= =∫ ∫ . Because ( )2

nψ is an even function of x for

all n, the product ( )2

nx ψ is an odd function of x. Therefore, x = 0 for all n. P7.7) Evaluate the average linear momentum of the quantum harmonic oscillator, px , for the ground state (n = 0) and first two excited states (n = 1 and n = 2). Use the hint about evaluating integrals in Problem P7.3.

2

2

210

12

0

1 3 5 (2 1)We use the standard integrals and2

4

n a xn n

ax

nx e d xa a

e dxa

π

π

∞ • • • • •−+

∞−

−=

⎛ ⎞= ⎜ ⎟⎝ ⎠

( )

( )

2 2

2

*

1 4 1 41 12 2

1 2

for 0,

x n n

x x

x

xx

dp x i d xd x

dn p e i e d xd x

p i x e d x

α α

α

ψ ψ

α απ π

α απ

−∞

− −∞

−∞

∞ −

−∞

⎛ ⎞= −⎜ ⎟

⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞= = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞= − −⎜ ⎟⎝ ⎠

h

h

h

Because the integrand is an odd function of , 0 for 0.xx p n= =

( ) ( )

2 2

2

1 4 1 41 13 32 2

1 232

4 4for 1,

4 1

x x

x

xx

dn p x e i x e d xd x

p i x x e d x

α α

α

α απ π

α απ

− −∞

−∞

∞ −

−∞

⎛ ⎞ ⎛ ⎞⎛ ⎞= = −⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠

⎛ ⎞= − −⎜ ⎟⎝ ⎠

h

h

Because the integrand is an odd function of , 0 for 1.xx p n= =

( ) ( )

( ) ( ) ( )

( ) ( )

2 2

2

2

1 4 1 41 12 22 2

1 22 2 3

1 23 5 2 3

for 2, 2 1 2 14 4

2 1 2 44

4 12 54

x x

x

xx

xx

dn p x e i x e d xd x

p i x e x x x d x

p i e x x x d x

α α

α

α

α αα απ π

α α α α απ

α α α απ

− −∞

−∞

∞ −

−∞

∞ −

−∞

⎛ ⎞⎛ ⎞ ⎛ ⎞= = − − −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞= − − − + +⎜ ⎟⎝ ⎠

⎛ ⎞= − − + −⎜ ⎟⎝ ⎠

h

h

h

Because the integrand is an odd function of , 0 for 3.xx p n= =

The result is general. 0 for all values of .xp n=

Page 57: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-8

P7.8) Evaluate the average of the square of the vibrational amplitude of the quantum harmonic oscillator about its equilibrium value, 2 ,x for the ground state (n = 0) and first two excited states (n = 1 and n = 2). Use the hint about evaluating integrals in Problem P7.3.

2

2

210

12

0

1 3 5 (2 1)We use the standard integrals and2

4

n a xn n

ax

nx e d xa a

e dxa

π

π

∞ • • • • •−+

∞−

−=

⎛ ⎞= ⎜ ⎟⎝ ⎠

( )( )

( )2 2

2 2

2 * 2

1 4 1 41 12 22 2

1 2 1 22 2 2

0

for 0,

2

n n

x x

x x

x x x d x

n x e x e d x

x x e d x x e d x

α α

α α

ψ ψ

α απ π

α απ π

−∞

− −∞

−∞

∞ ∞− −

−∞

=

⎛ ⎞ ⎛ ⎞= = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫ ∫

xk

21 2

22 12

12 2

= FHGIKJ = =

απ α

πα α µ

h

( )2 2

2 2

1 4 1 41 13 32 22 2

1 2 1 23 32 4 4

0

1 232

3 2

4 4for 1,

4 42

4 3 3 322 2 2

x x

x x

n x x e x x e d x

x x e d x x e d x

xk

α α

α α

α απ π

α απ π

α ππ α α α µ

− −∞

−∞

∞ ∞− −

−∞

⎛ ⎞ ⎛ ⎞= = ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

∫ ∫

h

( ) ( ) ( )

( ) ( )

2 2

2 2

1 4 1 41 12 2 2 22 2

1 2 1 22 2 6 4 2 2 6 4 2

0

for 2, 2 1 2 14 4

4 4 2 4 44 4

x x

x x

n x x e x x e d x

x x x x e d x x x x e d x

α α

α α

α αα απ π

α αα α α απ π

− −∞

−∞

∞ ∞− −

−∞

⎛ ⎞ ⎛ ⎞= = − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= − + = − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫ ∫

xk

21 2

24 3 3 2 22

44 15

24 3

21

25

25

2= FHG

IKJ − +FHG

IKJ = =

απ

αα

πα

αα

πα

πα α µ

h

P7.9) Evaluate the average of the square of the linear momentum of the quantum harmonic oscillator 2 ,xp for the ground state (n = 0) and first two excited states (n = 1 and n = 2). Use the hint about evaluating integrals in Problem P7.3.

Page 58: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-9

2

2

210

12

0

1 3 5 (2 1)We use the standard integrals and2

4

n a xn n

ax

nx e d xa a

e dxa

π

π

∞ • • • • •−+

∞−

−=

⎛ ⎞= ⎜ ⎟⎝ ⎠

( ) ( )

( )

2 2

2

22 * 2

0 02

1 4 1 41 1222 2

2

1 22 2

0

For = 0

x

x x

x

dp x x dxd x

n

de e dxd x

e x dx

α α

α

ψ ψ

α απ π

α α απ

∞− −

−∞

∞−

⎛ ⎞= −⎜ ⎟

⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞= − −⎜ ⎟⎝ ⎠

h

h

h

= − FHGIKJ −FHG

IKJ = =h h h2

1 22

4 2 212

απ

α πα

α πα

α µk

( ) ( )

( )

2 2

2

22 * 2

1 12

1 4 1 41 13 2 322 2

2

1 232 2 4 2

For = 1

4 4

4 3

x

x x

x

n

p x x dxx

xe xe dxx

x x e dx

α α

α

ψ ψ

α απ π

α α απ

∞− −

−∞

∞−

−∞

⎛ ⎞∂= −⎜ ⎟∂⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂= −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞= − −⎜ ⎟

⎝ ⎠

h

h

h

= −FHGIKJ −

= −FHGIKJLNM

OQP−LNM

OQP

FHG

IKJ

= =

−∞

∞zh

h

h h

23 1 2

2 4 2

23 1 2

22

2

4 3

4 34

3 12

32

32

2απ

α α

απ

α πα α

α πα α

α µ

αx x e dx

k

xc h

Page 59: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-10

( ) ( )

( ) ( )

( ) ( )

2 2

2

2

2

1 4 1 41 122 2 2 22 2

2

1 22 2 2 4 2

1 22 2 2 4 2

0

1 22

0

For = 2,

2 1 2 14 4

2 1 2 11 54

2 2 1 2 11 54

2 44

x x

x

x

x

x

n

p x e x e dxx

x e x x dx

x e x x dx

e

α α

α

α

α

α αα απ π

α α α α απ

α α α α απ

α απ

∞− −

−∞

∞−

−∞

∞−

∞−

⎛ ⎞∂⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎜ ⎟∂⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞= − − − +⎜ ⎟⎝ ⎠

⎛ ⎞= − − − +⎜ ⎟⎝ ⎠

⎛ ⎞= − ⎜ ⎟⎝ ⎠

h

h

h

h ( )4 6 3 4 2 224 21 5x x x dxα α α− + −

1 22 3 2

4 3 3 2 2

2

15 3 21 12 4 24 54 2 2 2 2

5 52 2

k

α π π π πα α απ α α α α α α α

α µ

⎛ ⎞⎛ ⎞= − − + −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= =

h

h h

P7.10) Using your results for Problems P7.6 through P7.9, calculate the uncertainties in the position and momentum σ σp xp p x x2 2 2 2 2 2= − = − and for the ground state (n = 0) and first two excited states (n = 1 and n = 2) of the quantum harmonic oscillator. Compare your results with the predictions of the Heisenberg uncertainty principle.

2 22 2 2 2

2

2

2 2

and

1 1For = 0, 02 2

02 2

1 12 2

p x

p

x

p x

p p x x

n k k

k k

p x

σ σ

σ µ µ

σµ µ

σ σ

= − = −

= − =

= − =

∆ ∆ = = ≥

h h

h h

h h

2

2

2 2

3 3For = 1, 02 2

3 302 2

3 12 2

p

x

p x

n k k

k k

p x

σ µ µ

σµ µ

σ σ

= − =

= − =

∆ ∆ = = ≥

h h

h h

h h

Page 60: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-11

2

2

2 2

5 5For = 2, 02 2

5 502 2

5 12 2

p

x

p x

n k k

k k

p x

σ µ µ

σµ µ

σ σ

= − =

= − =

∆ ∆ = = ≥

h h

h h

h h

P7.11) The vibrational frequency of 1H35Cl is 8.963 x 1013 s–1. Calculate the force constant of the molecule. How large a mass would be required to stretch a classical spring with this force constant by 1.00 cm? Use the gravitational acceleration on Earth at sea level for this problem.

( )

2 2

27 22 13 1

2

2 2

2

1 ; 42

1.008 34.969 1.661 10 kg4 amu 8.963 10 s1.008 34.969 amu

516 kg s

516 kg s 10 m 0.525 kg9.81 m s

k k

k

kF kx mg

kxmg

ν π µνπ µ

π−

− −

= =

× ×= × × × × ×

+== =

×= = =

P7.12) Two 1.00-g masses are attached by a spring with a force constant k = 500 kg s–2. Calculate the zero point energy of the system and compare it with the thermal energy kT. If the zero point energy were converted to translational energy, what would be the speed of the masses?

31 2

1 2

34 232

0 3

32110

23 1

2 32

3215 1

3

0.500 10 kg2

1.055 10 J s 500 kg s 5.28 10 J2 2 0.500 10 kg

5.28 10 J 1.27 10300 K 1.381 10 J K

1 v 5.28 10 J2

2 5.28 10 Jv = 7.27 10 m s2.00 10 kg

m m mm m

kE

EkT

m

µ

µ

− −−

−−

− −

−− −

= = = ×+

×= = = ×

×

×= = ×

× ×

= ×

× ×= ×

×

h

P7.13) Use x2 as calculated in Problem P7.8 as a measure of the vibrational

amplitude for a molecule. What fraction is x2 of the 127-pm bond length of the HCl

Page 61: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-12

molecule for n = 0, 1, and 2? The force constant for the 1H35Cl molecule is 516 N m–1. Use your results from Problem P7.8 in this problem. For n = 0,

121

3422 12

1 27 1

2 122

12

1.055 10 J s 7.59 10 m2 1.0078 34.96882 516 N m 1.66 10 kg amu

1.0078 + 34.9688

7.59 10 m = 5.97 10bond length 127 10 m

xk

x

µ

−−

− − −

−−

⎛ ⎞⎜ ⎟⎛ ⎞ ×⎜ ⎟= = = ×⎜ ⎟⎜ ⎟ ⎜ ⎟×⎝ ⎠ × × × ×⎜ ⎟⎝ ⎠

×= ×

×

h

For n = 1

121

2 342 11

1 27 1

2 11

12

3 3 1.055 10 J s 1.31 10 m2 1.0078 34.96882 516 N m 1.66 10 kgamu

1.0078 + 34.9688

1.31 10 m 0.103bond length 127 10 m

xk

x

µ

−−

− − −

⎛ ⎞⎜ ⎟⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎜ ⎟⎝ ⎠

× ×= = = ××× × × ×

×= =×

h

For n = 2

121

2 342 11

1 27 1

2 11

12

5 5 1.055 10 J s 1.70 10 m2 1.0078 34.96882 516 N m 1.66 10 kgamu

1.0078 + 34.9688

1.70 10 m 0.134bond length 127 10 m

xk

x

µ

−−

− − −

⎛ ⎞⎜ ⎟⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎜ ⎟⎝ ⎠

× ×= = = ××× × × ×

×= =×

h

P7.14) 1H35Cl has a force constant k = N m–1 and a moment of inertia of 2.644× 10–47 kg m2. Calculate the frequency of the light corresponding to the lowest energy pure vibrational and pure rotational transitions. In what regions of the electromagnetic spectrum do the transitions lie?

Page 62: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-13

213 1

271 1 516 kg s 8.963 10 s

1.008 34.969 1.661 10 kg2 2 amu1.008 34.969 amu

This frequency lies in the infrared region of the eletromagnetic spectrum. The lowest energy transition is = 0 = 1.

k

J J

νπ µ π

−−

−= = = ×× ×

×+

∆ ( ) ( )2342 222

2 2 47 20 0

11 1

1.055 10 J s2 1 4.20 10 J

2 2.644 10 kg m

6.35 10 s

This frequency lies in the microwave region of the eletromagnetic spectrum.

rot

rot

E Jr r

Eh

µ µ

ν

−−

×= + = = = ×

×∆

= = ×

h h

P7.15) The vibrational frequency for N2 expressed in wave numbers is 2358 cm–1. What is the force constant associated with the N ≡ N triple bond? How much would a classical spring with this force constant be elongated if a mass of 1.00 kg were attached to it? Use the gravitational acceleration on Earth at sea level for this problem.

( )

( )

2

27210 1 1

1

12

so 2

14.01amu 14.01amu 1.661 10 kg2 2.998 x10 cm s 2358cm2 14.01amu amu

2299 N m

kc

k c

k

k

ν νπ µ

π ν µ

π−

− −

= =

=

× ×= × × × ×

×

=

%

%

23

1

1.00 kg 9.81ms = 4.28 10 m2299 N m

F mgxk k

−−

×= = = ×

P7.16) The force constant for the 1H35Cl molecule is 516 N m–1. Calculate the vibrational zero point energy of this molecule. If this amount of energy were converted to translational energy, how fast would the molecule be moving? Compare this speed to the

root mean square speed from the kinetic gas theory, 3rms

kTm

=v for T = 300 K.

134 20

27 1

1 1 516 N m1.055 10 J s = 2.97 10 J1.0078 34.96882 2 1.66 10 kg amu1.0078 + 34.9688

kEµ

−− −

− −= = × × × ×

×× ×

h

The speed if converted to translational energy would be

( )20

127 1

2 2 2.97 10 Jv = 997 ms1.0078 + 34.9688 1.66 x10 kg amu

Em

−−

− −

× ×= =

×

The average speed from the kinetic gas theory is

Page 63: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-14

( )23 1

1rms 27 1

1

1rms

3 3 1.381 10 J K 300 Kv = 456 ms1.0078 + 34.9688 1.66 10 kg amu

v 997 ms 2.19v 456 ms

kTm

− −−

− −

× × ×= =

× ×

= =

P7.17) A gas-phase 1H19F molecule, with a bond length of 91.7 pm, rotates in a three-dimensional space. a) Calculate the zero point energy associated with this rotation. b) What is the smallest quantum of energy that can be absorbed by this molecule in a rotational excitation? a) There is no zero point energy because the rotation is not constrained. b) The smallest energy that can be absorbed is

( )( )

2342 2

227 1 12

22

2 1.055 x10 J s( 1) 1(1 1) 1.0078 18.99842 2 2 1.66 10 kg amu 91.7 x10 m

1.0078 + 18.99848.33 10 J

E J JI I

E

− − −

×= + = + =

×× × × ×

= ×

h h

P7.18) A 1H19F molecule, with a bond length 91.7 pm, absorbed on a surface rotates in two dimensions. a) Calculate the zero point energy associated with this rotation. b) What is the smallest quantum of energy that can be absorbed by this molecule in a rotational excitation? a) There is no zero point energy because the rotation is not constrained. b) The smallest energy that can be absorbed is

( )( )

2342 2 2

227 1 12

22

1.055 10 J s1.0078 18.99842 2 2 1.66 10 kg amu 91.7 10 m1.0078 + 18.9984

4.17 10 J

lmEI I

E

− − −

×= = =

×× × × × ×

= ×

h h

P7.19) The moment of inertia of 1H35Cl is 2.644× 10–47 kg m2. Calculate rotEkT

for

J = 0, 5, 10, and 20 at 298 K. For which of these values of J is 1rotEkT

≈ ?

Page 64: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-15

( )2342 222

47 2

0

22 215

215

23 1

2210

1.055 10 J s( 1) 1(1 1) ( 1) 2.089 10 ( 1)

2 2 2 2.664 10 kg m0

30 2.089 10 J = 6.267 10 J

6.267 10 J 1.511.381 10 J K 298 K110 2.089 10 J = 2.297 10

J

J

J

J

E J J J J J JI I

E

E

EkT

E

−−

=

− −=

−=

− −

−=

×= + = + = + = × +

× ×=

= × × ×

×= =

× ×

= × × ×

h h

20

2010

23 1

22 2020

2020

23 1

J

2.528 10 J 5.551.381 10 J K 298 K20 21 2.089 10 J = 8.774 10 J

8.774 10 J 21.21.381 10 J K 298 K

J

J

J

EkT

EEkT

−=

− −

− −=

−=

− −

×= =

× ×

= × × × ×

×= =

× ×

1rotEkT

≈ for J = 5.

P7.20) Using the Boltzmann distribution, calculate

0

Jnn

for 1H35Cl for the J values of

Problem P7.19 at T = 298 K. Does 0

Jnn

go through a maximum as J increases? If so, what

can you say about the value of J corresponding to the maximum?

( ) ( ) ( ) [ ]

( ) ( )

( ) ( )

( ) ( )

0

0

0

0

21 23 15

0

20 23 110

0

20 23 120

0

2 1 2 1 exp

1

2 5 1 exp 6.267 10 J 1.381 10 J K 298 K 2.42

2 10 1 exp 2.297 10 J 1.381 10 J K 298 K 0.082

2 20 1 exp 8.774 10 J 1.381 10 J K 2

JE E kTJJ

n J e J E kTnnnnnnnnn

− −

− − −

− − −

− − −

= + = + −

=

⎡ ⎤= × + − × × × =⎣ ⎦

⎡ ⎤= × + − × × × =⎣ ⎦

= × + − × × × 898 K 2.60 10−⎡ ⎤ = ×⎣ ⎦

0

Jnn

goes through a maximum because it has a value greater than one for J = 5. You can

only conclude that Jmax < 10.

Page 65: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-16

P7.21) By substituting in the Schrödinger equation for rotation in three dimensions,

show that the rotational wave function ( )1/ 2

25 3cos 116

θπ

⎛ ⎞ −⎜ ⎟⎝ ⎠

is an eigenfunction of the

total energy operator. Determine the energy eigenvalue.

( ) ( ) ( )

( ) ( )

22

2 2 20

1/ 2 1/ 22 2 2

2

2 2 20

2

20

, ,1 1sin ,2 sin sin

5 53cos 1 3cos 116 161 1sin

2 sin sin

52 16

Y YEY

r

r

r

θ φ θ φθ θ φ

µ θ θ θ θ φ

θ θπ π

θµ θ θ θ θ φ

µ π

⎡ ⎤∂ ∂⎛ ⎞∂− + =⎢ ⎥⎜ ⎟∂ ∂ ∂⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎛ ⎞ ⎛ ⎞∂ − ∂ −⎢ ⎥⎜ ⎟⎨ ⎬ ⎨ ⎬⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥∂ ⎜ ⎪ ⎪ ⎟ ⎪ ⎪⎩ ⎭ ⎩ ⎭− +⎢ ⎥⎜ ⎟∂ ∂ ∂⎢ ⎥⎜ ⎟

⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎛= −

h

h

h ( )( ) ( )

{ }( ) ( ) ( )

1/ 2 1/ 222 3 2

20

1/ 2 1/ 2 1/ 22 2 22 2 2 2

2 2 20 0 0

1 5 16cos sin 6sin 12 cos sinsin 2 16 sin

5 5 6 56 1 cos 12 cos 6 18cos 3cos 12 16 2 16 2 16

The energy

r

r r r

θ θ θ θ θθ θ µ π θ

θ θ θµ π µ π µ π

∂⎞ ⎡ ⎤ ⎛ ⎞ ⎡ ⎤− = − −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥∂⎝ ⎠ ⎣ ⎦ ⎝ ⎠ ⎣ ⎦

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎡ ⎤ ⎡ ⎤= − − − = − − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎣ ⎦⎣ ⎦⎝ ⎠ ⎝ ⎠ ⎝ ⎠

h

h h h

2 2

20

3 eigenvalue is , corresponding to ( 1) with = 2.2lE l l l

r Iµ= +

h h

P7.22) Show by carrying out the necessary integration that the eigenfunctions of the

Schrödinger equation for rotation in two dimensions, 1 1 and , 2 2

l lim inl le e m nφ φ

π π≠ are

orthogonal. 12

2 1 2 0

2

0

2

0

2

πφ

πφ φ φ

ππ π

φ φπ πFHGIKJ = − + −

= − − + − =

−z ze e d m n i m n d

m n i m n

m n

im inl l l l

l l l l

l l

l l 12

12

because is an integer

cos sin

cos sin

b g b gc h

b g b gc hb g

P7.23) In this problem you will derive the commutator ˆ ˆ ˆ, .x y zl l i l⎡ ⎤ =⎣ ⎦ h

a) The angular momentum vector in three dimensions has the form x y zl l l= + +l i j k where the unit vectors in the x, y, and z directions are denoted by , , and i j k. Determine lx, ly, and lz by expanding the 3 × 3 cross product l r p.= × The vectors

and r p are given by r kx y z= + +i j and .x y zp p p= + +p i j k b) Substitute the operators for position and momentum in your expressions for lx and ly. Always write the position operator to the left of the momentum operator in a simple product of the two. c) Show that ˆ ˆ ˆ, .x y zl l i l⎡ ⎤ =⎣ ⎦ h

Page 66: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-17

a)

( ) ( ) ( )( ) ( ) ( )

y z x yx zx y z

z y z x y x

z y x z y x

y z x yx zx y z

p p p pp pp p p

y p z p x p z p x p y p

y p z p z p x p x p y p

= = −

− − − + −

= − + − + −

i j kl i j + k

= i j k

i j k

b)

ˆ

ˆ ˆ ˆx y z

l i y i z i z i x i x i yz y x z y x

l i y z l i z x l i x yz y x z y x

⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂= − + + − + + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂= − − = − − = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠

i j kh h h h h h

h h h

c)

2 2

2 2 2 2

2

ˆ ˆ ˆ ˆ ˆ ˆ,x y x y y xl l l l l l

f f f fy z x z x z y zz y z x z x z y

f f f f f f f fy x z z x z x y z z y zz z x y z x z z y x z y

y x

⎡ ⎤ = − =⎣ ⎦⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

− − − + − −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

= − − + − + − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠∂

= −

h h

h h h h

h2 2 2 2

2 2 2 2 22

2 2 2 22 2 2 2 2 2

2

2 2

2

ˆ ˆ ˆ,x y z

f f f f fy yz zx zz x z x y z y x

f f f f fxy xz x yz zz y z y z x y x

f fy xx y

l l x y i i y x i ly x x y

∂ ∂ ∂ ∂+ + + −

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂+ − − − + =

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂

−∂ ∂

⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂⎡ ⎤ = − − = × − =⎜ ⎟ ⎜ ⎟⎣ ⎦ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

h h h h

h h h h h

h h

h h h h

P7.24) For molecular rotation, the symbol J rather than l is used as the quantum number for angular momentum. A 1H35Cl molecule whose bond length and force constant are 127 pm and 516 N m–1, respectively, has the rotational quantum number l = 10 and vibrational quantum number n = 0. a) Calculate the rotational and vibrational energy of the molecule. Compare each of these energies with kT at 300 K.

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Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-18

b) Calculate the period for vibration and rotation. How many times does the molecule rotate during one vibrational period? a)

( ) ( )( )

2342

272 210

20

20

23 1

10 11 1.055 10 J s11.0078amu 34.9688amu 1.661 10 kg2 2 1.275 10 m1.0078amu 34.9688amu amu

2.55 10 J

2.55 10 J 6.151.381 10 J K 300 K

rot

rot

rot

J JE

r

E

EkT

µ

−−

− −

× × ×+= =

× ×× × × ×+

= ×

×= =

× ×

h

134

27

20

20

23 1

1 1 516 N m1.055 10 J s1.0078amu 34.9688amu 1.661 10 kg2 21.0078amu 34.9688amu amu

2.97 10 J

2.97 10 J 7.171.381 10 J K 300 K

vib

vib

vib

kE n

E

EkT

µ

−−

− −

⎛ ⎞= + = × × ×⎜ ⎟ × ×⎝ ⎠ ×+

= ×

×= =

× ×

h

b) Calculate the period for vibration and rotation. How many times does the molecule vibrate during one rotational period?

( )

2

13

20

27 210

1 2; 222 2

2

2= = 1.43 10 s2 2.55 10 J

1.0078amu 34.9688amu 1.661 10 kg 1.275 10 m1.0078amu 34.9688amu amu

1.0078amu 34.9688amu 1.1 1.0078amu 34.9688amu2 2

rotrot

rotrot

vib

E IT

TEI

Tk

πω ω πν

π πω

π

µπ πν

−−

= = =

= =

×× ×

× ×× × ×

+

××

+= = =

27

141

661 10 kgamu = 1.12 10 s

516 N m

−−

×

×

It vibrates 13

14

1.43 10 s 12.81.12 10 s

rot

vib

TT

×= =

×times in one rotational period.

P7.25) At what values of θ does ( ) ( )1/ 2

0 22

5, 3cos 116

Y θ φ θπ

⎛ ⎞= −⎜ ⎟⎝ ⎠

have nodes? Are the

nodes points, lines, planes, or other surfaces?

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Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-19

Y2

0 θ φ,b g has nodes when 3 12cos θ −c h has nodes. This occurs for 0.955 and 0.955 radians or 54.7 and 125.3 degrees.θ π= − These surfaces are cones.

P7.26) The wave functions px and dxz are linear combinations of the spherical harmonic functions, which are eigenfunctions of the operators ˆ ,H 2ˆ ,l and zl for rotation in three dimensions. The combinations have been chosen to yield real functions. Are these functions still eigenfunctions of ˆ ?zl Answer this question by applying the operator to the functions.

$ sin cos sin sinL p i iz x = −∂∂

=h hθ π

θ φπ

θ φ34

34

. This shows that px is not an

eigenfunction of ˆ .zL

$ sin cos cos sin cos sinL d i iz xz = −∂∂

=h hθ π

θ θ φπ

θ θ φ154

154

. This shows that dxz is

not an eigenfunction of ˆ .zL

P7.27) Show that the function ( ) ( )1/ 2

0 22

5, 3cos 116

Y θ φ θπ

⎛ ⎞= −⎜ ⎟⎝ ⎠

is normalized over the

interval 0 0 2≤ ≤ ≤ ≤θ π φ π and .

( )( ) ( )

( )

22 20 2

20 0

4 2

0

5, sin 3cos 116

52 sin 9cos 6cos 116

Y d d d

d

π π

π

θ φ τ φ θ θ θπ

π θ θ θ θπ

⎛ ⎞= −⎜ ⎟⎝ ⎠

⎛ ⎞= − +⎜ ⎟⎝ ⎠

∫∫ ∫ ∫

( )( ) ( )

( )

22 20 2

20 0

4 2

0

5 3

0

5, sin 3cos 116

52 sin 9cos 6cos 116

5 9 5 18cos 2cos cos 4 2 18 5 8 5

Y d d d

d

π π

π

π

θ φ τ φ θ θ θπ

π θ θ θ θπ

θ θ θ

⎛ ⎞= −⎜ ⎟⎝ ⎠

⎛ ⎞= − +⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= − + − = − + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

∫∫ ∫ ∫

P7.28) Is it possible to know simultaneously the angular orientation of a molecule rotating in a two dimensional space and its angular momentum? Answer this question by

evaluating the commutator , iφφ

⎡ ⎤∂−⎢ ⎥∂⎣ ⎦h .

Page 69: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-20

( ) ( ) ( ) ( ),

,

d fdfi f i i i f

d d

i i

φ φφφ φ φ φ φ

φ φ φ

φφ

⎡ ⎤⎡ ⎤∂ ⎣ ⎦− = − + =⎢ ⎥∂⎣ ⎦⎡ ⎤∂

− =⎢ ⎥∂⎣ ⎦

h h h h

h h

Because the commutator is not equal to zero, it is not possible to simultaneously know the angular orientation of a molecule rotating in a two-dimensional space and its angular momentum. P7.29) At 300 K, most molecules are in their ground vibrational state. Is this also true

for their rotational degree of freedom? Calculate 1 5

0 0

and J J

J J

n nn n

= =

= =

for the H35Cl molecule

whose bond length is 1.27× 10–10 m. Make sure that you take the degeneracy of the levels into account.

( )

( ) ( )

( )

( )

2

2

1-

21

0

234

227 1 10 23 1

1

0

1-

25

0

2 1 = e1

1.055 10 J s 1 1+13 exp 1.0078 34.96882 amu 1.66 10 kg amu 1.27 10 m 1.381 10 J K 300 K

1.0078 + 34.9688

= 2.708

2 1 = e1

J JI kTJ

J

J

J

J JI kTJ

J

n Jn

nn

n Jn

+

=

=

− − − − −

=

=

+

=

=

+

⎡ ⎤⎢ ⎥×

= −⎢ ⎥×⎢ ⎥× × × × × × × ×

⎢ ⎥⎣ ⎦

+

=

h

h

( ) ( )

( )

234

227 1 10 23 1

5

0

1.055 10 J s 5 5+111 exp 1.0078 34.96882 amu 1.66 x10 kg amu 1.27 10 m 1.381 10 JK 300 K

1.0078 +34.9688

= 2.369J

J

nn

− − − − −

=

=

⎡ ⎤⎢ ⎥×−⎢ ⎥

×⎢ ⎥× × × × × × ×⎢ ⎥⎣ ⎦

P7.30) Draw a picture (to scale) showing all angular momentum cones consistent with l = 5. Calculate the half angles for each of the cones.

Page 70: Physical Chemistry II

Chapter 7/A Quantum Mechanical Model for the Vibration and Rotation of Molecules

7-21

ml = +5

ml = +4

ml = +3

ml = +2

ml = +1

ml = 0

ml = -1

ml = -2

ml = -3

ml = -4

ml = -5

The half angles of the cones measured from the positive and negative z axis are

( )1

55cos 0.420 radians

5 5 1lmθ−

=

⎛ ⎞⎜ ⎟= =⎜ ⎟+⎝ ⎠

( )1

44cos 0.752 radians

5 5 1lmθ−

=

⎛ ⎞⎜ ⎟= =⎜ ⎟+⎝ ⎠

( )1

33cos 0.991radians

5 5 1lmθ−

=

⎛ ⎞⎜ ⎟= =⎜ ⎟+⎝ ⎠

( )1

22cos 1.20 radians

5 5 1lmθ−

=

⎛ ⎞⎜ ⎟= =⎜ ⎟+⎝ ⎠

( )1

11cos 1.39 radians

5 5 1lmθ−

=

⎛ ⎞⎜ ⎟= =⎜ ⎟+⎝ ⎠

( )1

00cos 1.57 radians

5 5 1lmθ−

=

⎛ ⎞⎜ ⎟= =⎜ ⎟+⎝ ⎠

as well as π minus these values for the negative values of ml.