1

CH107 Special Topics

Part A: The Bohr Model of the

Hydrogen Atom (First steps in

Quantization of the Atom)

Part B: Waves and Wave Equations

(the Electron as a Wave Form)

Part C: Particle (such as an Electron)

in a Box (Square Well) and Similar

Situations

2

A. The Bohr Model of the Hydrogen Atom

Photon

n3 n2

n1 n3

3

Setting a Goal for Part A

• You will learn how Bohr imposed Planck’s hypothesis on a classical description of the Rutherford model of the hydrogen atom and used this model to explain the emission and absorption spectra of hydrogen (Z = 1) and other single electron atomic species (Z = 2, 3, etc).

4

Objective for Part A

• Describe how Bohr applied the quantum hypothesis of Planck and classical physics to build a model for the hydrogen atom (and other single electron species) and how this can be used to explain and predict atomic line spectra.

5

Setting the Scene

At the turn of the 19th/20th centuries, classical

physics (Newtonian mechanics

and Maxwellian wave theory) were unable

explain a number of observations relating to

atomic phenomena:• Line spectra of atoms (absorption or emission)• Black body radiation• The photoelectric effect• The stability of the Rutherford (nuclear) atom

6

The Beginnings of Bohr’s Model

In 1913, Niels Bohr, a student of Ernest

Rutherford, put forward the idea of

superimposing the quantum principal of Max

Planck (1901) on the nuclear model of the H

atom.

He believed that exchange of energy in

quanta (E = h) could explain the lines in the

absorption and emission spectra of H and

other elements.

7

Basic Assumptions (Postulates) of Bohr’s Original Model of the H Atom

1. The electron moves in a circular path around the nucleus.

2. The energy of the electron can assume only certain quantized values.

3. Only orbits of angular momentum equal to integral multiples of h/2 are allowed (meru = nħ,

which is equation (1); n = 1, 2, 3, …)

4. The atom can absorb or emit electromagnetic radiation (h) only when the electron transfers between stable orbits.

8

First Steps: Balance of Forces

+

e- (me)

(Ze+)

ru

The centripetal force of circular motion balances the electrostatic force of attraction:

4 r20

Ze2=

meu2

ror meu2

=4 r0

Ze2

0 is known as the permitivity of vacuum

(2)

9

Determination of Potential Energy (PE or V) and Kinetic Energy (KE)

In general, V(r) =_

F(r) dr0

r

For the Bohr H atom, V(r) =_ Ze2

40 r

The PE arises from the attractive electrostatic forcebetween the electron and the nucleus.

The KE exists by virtue of the electron's (circular) motion.

KE =12

meu2

10

Determination of Total Energy and Calculation of u

Hence the total energy is12

meu2 Ze2

40 r_

=12

meu2_ (sinceZe2

40 r= meu2

) (3)

Imposition of the quantum restriction meru = nh (1)and substituting for r in equation (2) gives

u = Ze2

2h0n(4)

11

Determination of Total Energy

Substitution of the value of u from equation (4)into equation (2) gives

En =_ e4me Z2

n2802h2

(5)

This is the Bohr equation for quantized electronenergy levels: the integer n, introduced inequation (1) with the quantum hypothesis, definesthe energy for each stable orbit, known as energylevels.

12

Determination of Total Energy - Continued

The expression inside the brackets ofequation (5) is a constant, equivalentto 2.18 x 10-18 J.This is a convenient unit of electronicenergy and is called the rydberg.Equation (5) thus reduces to

En = _ Z2

n2(rydberg)

where n = 1, 2, 3,...and for H, Z = 1

(6)

13

Energy Level Diagram for the Bohr H Atom

n En0.00 J

-2.18 x 10-18 J_

1.00 rydberg)(

-5.45 x 10-19 J (_ 1

4rydberg )

-2.42 x 10-19 J (_ 1

9rydberg )

(_ 1

16rydberg )-1.36 x 10-19 J

1

2

3

4

14

Determination of Orbit Radius

Elimination of u between equations (1) and (4),and solving for r, gives

rn =0h

2

mee2(6)

This equation describes the radii of the allowed orbitscorresponding to different values of n, and with energiesgiven by equation (5).As before, Z = 1 for the H atom and n = 1, 2, 3,...

n2

Z

15

Determination of Orbit Radius - Continued

All the parameters in equation (6),apart from Z and n2,are constants, which is equivalent to5.29 x 10-11 m (0.529 A).This is known as the Bohr radius (a0),and is a convenient measure ofelectronic distance in atoms. Thusequation (6) reduces to

rn = n2

Za0 (7)

16

Electronic Radius Diagram of Bohr H Atom

n = 1n = 2

n = 3

n = 4

r4 =16a0

r3 =9a0

r2 =4a0

r1 = a0

Each electronic radius corresponds to an energy level with the same quantum number

17

The Line Spectra of Hydrogen

The major triumph of the Bohr model of the H atom was its ability to explain and predict the wavelength of the lines in the absorption and emission spectra of H, for the first time.Bohr postulated that the H spectra are obtained from transitions of the electron between stable energy levels, by absorbing or emitting a quantum of radiation.

18

Electronic Transitions and Spectra in the Bohr H atom

Photon

n3 n2

n1 n3

Absorption

Electron promotion

Electron relaxation

Emission

19

Qualitative Explanation of Emission Spectra of H in Different Regions of the Electromagnetic

Spectrum

En

erg

y

n = 1

2

3

4

5

7

6

Lyman series(ultraviolet)

Balmer series(visible)

Paschen series(near infrared)

Brackett series(mid infrared)

20

Quantitative Explanation of Line Spectra of Hydrogen and Hydrogen-Like Species

When H or H-like species (a one-electron ion)undergoes a transition between two energylevels Ei and Ef, E = + h_

ni

nf ni

nf

Emission Absorption

E h h

h =Z2e4me

802h2 nf

2

Z2e4me

802h2 ni

2

ni > nf

_h =

Z2e4me

802h2 ni

2

Z2e4me

802h2 nf

2

nf > ni

_

21

…….Continued

= Z2e4me

802h3 nf

2 ni2

1 1_

= Z2e4me

802h3 ni

2 nf2

1 1_

(Emission)

(Absorption)

(8)

(9)

From the previous slide,

802h3

e4meThe constant is equivalent to Rydberg's

empirical constant (3.29 x 1015 s-1), hence

= (3.29 x 1015 s-1) Z2nf

2 ni2

1 1_ (Emission) (10)

= (3.29 x 1015 s-1) Z2ni

2 nf2

1 1_ (Absorption) (11)

22

Conclusion

For the H atom (Z = 1), the predicted

emission spectrum associated with nf = 1

corresponds to the Lyman series of lines in

the ultraviolet region.

That associated with nf = 2 corresponds to

the Balmer series in the visible region, and

so on.

Likewise, the lines in all the absorption

spectra can be predicted by Bohr’s equations.

23

Aftermath – the Successes and Failures of Bohr’s Model

• For the first time, Bohr was able to give a

theoretical explanation of the stability of the

Rutherford H atom, and of the line spectra of

hydrogen and other single electron species

(e.g. He+, Li2+, etc).• However, Bohr’s theory failed totally with two-and

many-electron atoms, even after several drastic

modifications. Also, the imposition of quantization

on an otherwise classical description was uneasy.• Clearly a new theory was needed!

24

B. Waves and Wave Equations

25

Setting a Goal for Part B

• You will learn how to express equations for wave motions in both sine/cosine terms and second order derivative terms.

• You will learn how de Broglie’s matter wave hypothesis can be incorporated into a wave equation to give Schrödinger-type equations.

• You will learn qualitatively how the Schrödinger equation can be solved for the H atom and what the solutions mean.

26

Objective for Part B

•Describe wave forms in general and matter (or particle) waves in particular, and how the Schrödinger equation for a 1-dimensional particle can be constructed.•Describe how the Schrödinger equation can be applied to the H atom, and the meaning of the sensible solutions to this equation.

27

Waveamplitudey(x)

Basic wave properties

A = amplitude = wavelength = frequencyc = velocity

c = E = h= hc/

= hc =

= wavenumber

The Basic Wave Form

(1)

(2)

28

Basic Wave Equations

For a travelling wave,

y(x) = Asin 2x

_ 2t

For a standing wave,

y(x) = Asin 2x

(3)

(4)

29

Characteristics of a Travelling Wave

30

Characteristics of a Standing Wave

31

Wave Form Related to Vibration and Circular Motion

x

y

Lookingalong xaxis

The wave formappears as avibrationor oscillation

This can beresolved intotwo opposingcircular motions

32

The Dual Nature of Matter – de Broglie’s Matter Waves

• We have seen that Bohr’s model of the H atom could not be used on multi-electron atoms. Also, the theory was an uncomfortable mixture of classical and modern ideas.

• These (and other) problems forced scientists to look for alternative theories.

• The most important of the new theories was that of Louis de Broglie, who suggested all matter had wave-like character.

33

de Broglie’s Matter Waves

• De Broglie suggested that the wave-like character of matter could be expressed by the equation (5), for any object of mass m, moving with velocity v.

• Since kinetic energy (Ek = 1/2mv2) can be written as

• De Broglie’s matter wave expression can thus be

written

=h

mv(5)

Ek =(mv)2

2m

=h

2mEk(6)

34

de Broglie’s Matter Waves, Continued

• Since h is very small, the de Broglie wavelength will be too small to measure for high mass, fast objects, but not for very light objects. Thus the wave character is significant only for atomic particles such as electrons, neutrons and protons.

• De Broglie’s equation (5) can be derived from

(1) equations representing the energy of photons (from Einstein – E = mc2 – and Planck – E = hc/)

and also

(2) equations representing the electron in the Bohr H atom as a standing wave (mevr = nh/2; n = 2r)

35

The Electron in an Atom as a Standing Wave

An important suggestion of de Broglie was that the electron in the Bohr H atom could be considered as a circular standing wave

36

Differential Form of Wave Equations

• Consider a one-dimensional standing wave. If we suppose that the value y(x) of the wave form at any point x to be the wave function x, then we have, according to equation (4)

• Of particular interest is the curvature of the wave function; the way that the gradient of the gradient of the plot of versus x varies. This is the second derivative of with respect to x.

(x) = A sin2x

or B cos

2x

(7)

37

Differential Form of Wave Equations, Continued

Thus

Equation (9) is a second order differential equation

whose solutions are of the form given by equation

(7).

d2(x)

dx2= _ A 2

sin 2x

2(8)

= 2

2(x) (9)or d2(x)

dx2

_

38

Differential Wave Equation for a One-Dimensional de Broglie Particle Wave

We now consider the differential wave equation for

a one-dimensional particle with both kinetic energy

(Ek) and potential energy (V(x)).

If =h

mv, then

d2dx2 =

_ 2mv

h

2(x)

Manipulating the above equation to get a kineticenergy term, since Ek = (mv)2/2m

_ h2

82m

d2dx2

=(mv)2

2m(x)

_ h2

82m

d2dx2

= (x)Ek (10)

or,

39

The Schrödinger Equation for a Particle Moving in One Dimension

• Equation (10) shows the relationship between the second derivative of a wave function and the kinetic energy of the particle it represents.• If external forces are present (e.g. due to the presence of fixed charges, as in an atom), then a potential energy term V(x) must be added. • Since E(total) = Ek + V(x), substituting for Ek in Equation (10) gives

This is the Schrödinger equation for a particle moving in one dimension.

_ h2

82m

d2dx2 = (x)E+ V(x) (11)

40

The Schrödinger Equation for the Hydrogen Atom

Erwin Schrödinger (1926) was the first to act upon de

Broglie’s idea of the electron in a hydrogen atom

behaving as a standing wave. The resulting equation

(12) is analogous to equation (11);

It represents the wave form in three dimensions and is

thus a second-order partial differential equation. h2

82m

_

x2

2

y2

2

z2

2

+ + + V(x,y,z) (x,y,z) = E (x,y,z)

(12)

2 _ h2

2m+ V = Eor

41

The Schrödinger Equation, Continued

• The general solution of equations like equation

(12) had been determined in the 19th century (by

Laguerre and Legendre).• The equations are more easily solved if

expressed in terms of spherical polar coordinates

(r,), rather than in cartesian coordinates (x,y,z), in

which case, 2

in equation (12) becomes

1

r2

2

r2

rr +1

r2sin sin +

1

r2sin2 2

42

Erwin Schrödinger

Students:

I hope you are staying awake

while the professor

talks about my work!

Love,

Erwin

43

Spherical Polar Coordinates

z

x

y

r

0

P

is the angle between OPthe z axis.

(the azimuthal angle) isthe angle between theprojection of OP ontothe xy plane.

The electron position withrespect to the nucleusin spherical polar coordinatesis r,, (in cartesian coordinatesthis is x,y,z)

x = rsincosy = rsinsinz = rcosr2 = x2 + y2 + z2

44

Solutions of the Schrödinger Equation

for the Hydrogen Atom • The number of solutions to the Schrödinger

equation is infinite.• By assuming certain properties of (the wave

function) - boundary conditions relevant to the physical nature of the H atom - only solutions meaningful to the H atom are selected.

• These sensible solutions for (originally called specific quantum states, now orbitals) can be expressed as the product of a radial function [R(r)] and an angular function [Y(,)], both of which include integers, known as quantum numbers; n, l and m (or ml ).

45

Solutions of the Schrödinger Equation for the Hydrogen Atom, Continued

(r,,) = Rnl(r)Ylm(,) (13)

• The radial function R is a polynomial in r of degree n – 1 (highest power r(n-1), called a Laguerre polynomial) multiplied by an exponential function of the type e(-r/na0) or e(-/n), where a0 is the Bohr radius.

• The angular function Y consists of products of polynomials in sin and cos (called Legendre polynomials) multiplied by a complex exponential function of the type e(im).

46

Solutions of the Schrödinger Equation for the Hydrogen Atom, Continued

• The principal quantum number is n (like the Bohr quantum number = 1, 2, 3,…), whereas the other two quantum numbers both depend on n.

• l = 0 to n - 1 (in integral values).• m = -l through 0 to +l (again in integral values).• The energies of the specific quantum states (or

orbitals) depend only on n for the H atom (but not for many-electron atoms) and are numerically the same as those for the Bohr H atom.

47

Orbitals

• Orbitals where l = 0 are called ‘s orbitals’; those with l = 1 are known as ‘p orbitals’; and those with l = 2 are known as ‘d orbitals’.

• When n = 1, l = m = 0 only; there is only one 1s orbital.

• When n = 2, l can be 0 again (one 2s orbital), but l can also be 1, in which case m = -1, 0 or +1 (corresponding to three p orbitals).

• When n = 3, l can be 0 (one 3s orbital) and 1 (three 3p orbitals) again, but can also be 2, whence m can be –2, -1, 0, +1 or +2 (corresponding to five d orbitals).

48

Energy Levels of the H atom

49

The 1s Wave Function of H and Corresponding Pictorial Representation

The 1s wave function is the solution of the Schrödingerequation when n = 1; l= 0; m = 0. It represents the orbitalwith the lowest energy.

100 =1a0

-3/2 e-r/a2 2

Y00 is a constant forall cases where l = m = 0(s orbitals), indicatingspherical symmetry

R10 is a functionof r0 and of

The 3D boundarysurface of the 1sorbital

0

e-r/a0

50

The 2pz Wave Function of H and Corresponding Pictorial Representation

The 2pz wave function is the solution of the Schrödingerequation when n = 2; l = 1; m = 0. It represents the p orbitalpointing along the z axis.

210 =2

Y10 is a functionof only, indicatingdirectionality alongthe z axis

R21 afunction ofr, as well asof

1

2 6a0

-5/2 r e-r/2a0 3 cos

The 3D boundarysurface of the 2pzorbital

e-r/2a0

51

C. Particle in a One-Dimensional Box

V(x)

0x

a

oooo

0

52

Setting a Goal for Part C

• You will learn how the Schrödinger equation can be applied to one of the simplest problems; a particle in a one-dimensional box or energy well.

• You will learn how to calculate the energies of various quantum states associated with this system.

• You will learn how extend these ideas to three dimensions.

53

Objective for Part C

• Describe how the Schrödinger equation can be applied to a particle in a one-dimensional box (and similar situations) and how the energies of specific quantum states can be calculated.

54

Particle in a One-Dimensional Box

• The simplest model to which the Schrödinger equation can be applied is the particle (such as a ‘1-D electron’) in a one-dimensional box or potential energy well.

• The potential energy of the particle is 0 when it is in the box and beyond the boundaries of the box; clearly the particle is totally confined to the box.

• All its energy will thus be kinetic energy.

55

Defining the Problem

V(x)

0x

L

oooo

0

V(x) = 0 (0 < x < L)

V(x) = (x < 0 or x > L)oo

56

Setting up the Schrödinger Equation

82m

_ h2 d2E

dx2=

The Schrödinger equation for this modelis derived from equation (11) in Part B, withV(x) = 0 for 0 < x < L

(1)

Separating variables in the Schrödingerequation gives equation (2)

h2

82mEd2

dx2= _ (2)-k2

57

Solution of the Schrödinger Equation and use of the First Boundary

Condition

The solution to this equation is(x) = A sin(kx) or B cos(kx),but after imposition of the boundarycondit ion that (x) = 0 at x = 0,the solution must be

(x) = A sin(kx)

Substituting this into equation 1 gives

_ A k2sin(kx) = A sin(kx)h2

82mE_ (3)

58

Evaluation of the Constant k and Use of the 2nd Boundary Condition

From equation (3),

k =

12

and hence

(x) = A sin x

h2

82mE

12

h2

82mE (4)

Imposition of the 2nd boundary condit ion,that (x) = 0 at x = L, implies equation (5)

12

L = n

since sin(x) is zero only when x = n

(5),h2

82mE

59

Determination of the Energy LevelsThis boundary condition issatisfied if E is restricted tovalues that satisfy equation(5): that is if En is the valueof energy that satisfiesequation (5) for given allowedvalue of n, then

En =n2h2

8mL2

(n = 1, 2, 3,....)

(6)

This gives rise to the setof energy levels opposite,showing the correspondingwave forms

The spacing of energy levels;

h2

8 mL2(2n + 1)_E = En-1 En =

60

Determination of the Constant A

From equations (4) and (5), it can be seen that

nxL

(x) = A sin (7)

A is called the normalization constant . To evaluate A,we refer to the well-behaved nature of that theintegral of 2 over all space (here between x = 0 andx = L) must be 1, since the particle is somewhere inthe box.

i.e

0

a2dx =

0

anxL

A2sin2 = 1

from which, A = 2L

Hence (x) = sinn x

( n = 1, 2, 3,....)L

(8)2L

dx

or A2 L2 = 1

61

A Particle in a Three-Dimensional Box

• The arguments in the previous slides can be extended to a particle confined in a 3D box of lengths Lx, Ly and Lz.

• Within the box, V(x,y,z) = 0; outside the cube it is

• A quantum number is needed for each dimension

and the Schrödinger equation includes derivatives with respect to each coordinate.

• The allowed energies for the particle are given by

Enxnynz=

h2

8mnx

2

Lx2 + +

ny2

Ly2

nz2

Lz2

(9)

62

Calculation of Energies of a Particle in a 3-D Box

Comparative energies of a particle confinedto a box of sides 2L, L, L.

En n nx y z=

h2

8m

nx2

(2L)2ny

2

L2

nz2

L2+ +

For the ground state, nx = ny = nz = 1

Hence E111 =h2

8mL2

14

1 1+ + =9h2

32mL2

63

….Continued

E211

For the excited states where one of nx, ny or nz is 2,the others are 1, we have

=h2

8mL2

4

4+ 1 + 1 =

3h2

8mL2

12h2

32mL2

E121 = E112h2

8mL2=1

4+ 4 + 1 =

21h2

32mL2

The last two are degenerate (of the same energy).

Note that there are many other excited states; E311,E321, etc.

If all the box sides had been L, then all three of theabove excited state energies would be degenerate.

64

Calculation of Energy Spacing in Different Situations

Consider two situations:

(1) an electron in a one-dimensional box of length 1.0 A.

(2) an electron in a cube of lengths 10 cm on an edge.

Calculate the energy difference between the ground stateand the first excited state.

(1) E =3h2

8mL2=

3 (6.626 x 10-34 Js)2

8 (9.11 x 10-31 kg)(1.0 x 10-10 m)2

= 1.807 x 10-17 J ( 10,880 kJ/mol)

65

…Continued

(2) For the cube, where Lx = Ly = Lz = L = 10 cm (0.1 m)

E111 (ground state) = h2

8mL212 + 12 + 12

E211 = E121 = E112 =h2

8mL222 + 12 + 12

E =h2

8mL2

3

(First excited state)

E = 3 (6.626 x 10-34 Js)2

8 (9.11 x 10-31 kg)(0.1 m)2= 1.807 x 10-35 J

This is equivalent to 1.088 x 10-14 kJ/mol. The energy levelsin general are so close together that they appear continuous;quantum effects are minimal, quite unlike the case with theelectron in a 1-dimensional box of atom-size length 1.0 A.