i be qp makeup exam 15-07-2011 ns scheme

7/31/2019 i Be Qp Makeup Exam 15-07-2011 Ns Scheme

1/141/14

Reg.No.

MANIPAL INSTITUTE OF TECHNOLOGYMANIPAL UNIVERSITY, MANIPAL

I YEAR B.E. END-SEMESTER MAKE- UP EXAMINATION JULY 2011SUBJECT: ENGINEERING PHYSICS (PHY101/102)

(NEW SYLLABUS)

Time: 3 Hrs. Max. Marks: 50

Note: Answer any FIVE FULL questions. Each question carries 10 marks

Answer all the sub questions of a main question in a continuous sequence.Write specific and precise answers.Write question number within the margin. Draw neat sketches wherever necessary.

Physical Constants:

Speed of light in vacuum = 3.00 X 108

m/s Electron charge = 1.60 X 1019

C

Mass of the alpha particle, m = 6.64 X 1027

kg 360 = 2 rad

Boltzmann constant = 1.38 X 1023

J/ K Plancks constant = 6.63 X 1034

J.s1 atomic mass unit, u = 1.66 X 10

27kg; Mass of Carbon = 12 u; Mass of Oxygen = 16 u

Ionization energy for Hydrogen in the ground state = 13.6 eV Bohr radius = 52.9 pm

1A Explain a wavepacket, and its (i) group speed (ii) phase speed. 3 M

1B Which are the features of photoelectric effect-experiment explained by Einsteinsphotoelectric equation? 2 M

1C Obtain an expression for the radius of mTH order bright ring in the case ofNewtons rings. 5 M

2A Explain Rayleighs criterion for resolving images due to a circular aperture. 2 M

2B Obtain an expression for the quantized rotational energy of a diatomic molecule. 3 M

2C Sketch the lowest three energy states, wave-functions, probability densities for the

particle in a one-dimensional infinite potential well. 3 M

2D Explain (i) spontaneous emission (ii) stimulated emission with reference to lasers. 2 M

3A Give the physical interpretation of spin magnetic quantum number, ms. 3 M

3B Assuming the expression for the allowed states of a particle in a 3- dimensional box,

derive the density-of-states function. 5 M

3C Explain Meissner effectin superconductors. 2 M

4A Monochromatic light with wavelength 538 nm falls on a slit with width 25.2 m. The

distance from the slit to a screen is 3.48 m. Consider a point on the screen 1.13 cmfrom the central maximum. Calculate the ratio of the intensity at this point to the

intensity at the central maximum. 3 M


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MIT- MU I BE SE COND SEMESTER - MAKE- UP EXAMINATION JULY 2011 ENGINEERING PHYSICS (NS)

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4B How many complete fringes appear between the first minima of the fringe envelope on

either side of the central maximum for a double-slit pattern if = 557 nm, d = 0.150 mm,and a = 0.030 mm? 2 M

4C Electrons bombard a molybdenum target, producing both continuous and

characteristic x-rays. If the accelerating potential applied to the x-ray tube is50.0 kV, what values of (i) MIN (ii) K (iii) K result ? The energies of the

K-shell, L-shell and M-shell in the molybdenum atom are 20.0 keV, 2.6 keV,

and 0.4 keV respectively. 3 M

4D When red light in vacuum is incident at the polarizing angle on a certain glass slab, the

angle of refraction is 31.8. What are (i) the index of refraction of the glass and (ii) thepolarizing angle? 2 M

5A A grating has 40000 ruling spread over76 mm. What is its expected dispersion D in /nm for sodium light of wavelength 589 nm in the first order? 3 M

5B An alpha particle in a nucleus can be modeled as a particle moving in a box of length

1.0 x 1014 m. Using this model, estimate the wavelength and momentum of the alpha

particle in its first excited energy state (n=2). 2 M

5C The frequency of photon that causes v = 0 to v = 1 transition in the CO

molecule is 6.42 x 1013 Hz. Ignore any changes in the rotational energy. (i)

Calculate the force constant, k for this molecule. (ii) What is the maximum classicalamplitude of vibration for this molecule in the v = 0 vibrational state? 3 M

5D The lifetime of an electron in the state n = 2 in hydrogen is 10 ns. What is the

uncertainty in electron energy in this state? 1 M

6A Calculate the probability of finding the electron in the ground state of an infinite

potential well of length L, between x = 0 and x = L /4. 3 M

6B A hydrogen atom is in the first excited state (n = 2). Using the Bohr theory ofthe atom, calculate (i) the radius of the orbit (ii) the kinetic energy of the

electron (iii) the potential energy of the system and (iv) the total energy of the

system. 3 M

6C Most solar radiation has a wavelength of 1 m. What energy gap should thematerial in solar cell have in order to absorb this radiation ? 2 M

6D Calculate the energy of a conduction electron in silver at 800 K, assuming theprobability of finding an electron in that state is 0.950. The Fermi energy is

5.48 eV at this temperature. 2 M


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PHYSICS DEPARTMENT, MIT, MANIPALSCHEME OF EVALUATION

II SEMESTER MAKE- UP EXAMINATION- PHY 101/102 on 15-07-2011

Q1A Explain a wavepacket, and its (i) group speed (ii) phase speed. 3 M

Ans. 1A Quantum particle which is associated with the dual nature is represented with thehelp of a wavepacket.To create a wave packet, large number of simple harmonic waves are added. This additioncreates a small localized region of space where constructive interference takes place and iscalled a wavepacket, which represents the particle.

------------------------ 1 M

(i) The amplitude envelope of the wavepacket can travel through space with a different

speed than the individual waves. This speed is called the group speed or the speed of

the wave packet. ---------------------- M

For a superposition of large number of waves to form a wave packet, the group speed isgiven by

---------------------- M

Here d and dk are difference in angular frequencies and wavenumbers of the addedwaves respectively.

(ii) The phase speed is the speed with which the resultant wave crest moves ------ M

Is the phase speed of the wavepacket. ---------------------- M

Q1B Which are the features of photoelectric effect-experiment explained by Einsteinsphotoelectric equation? 2 M

Ans. 1B Einsteins photoelectric equation showed that (i) the maximum kinetic energy ofthe emitted electrons, Kmax depends only on frequency of the incident light.

(ii) the almost instantaneous emission of photoelectrons due to one -to one interactionbetween photons and electrons.

(iii) the necessity of a minimum light frequency, thereby a minimum photon energy forthe photoelectric effect to take place since photons should have energy greater

than the work function in order to eject an electron.

(iv) the existence of a cutoff frequency fc which is related to by fc = /h. If theincident frequency f is less than fc , no emission of photoelectrons.

----------------------( 4) = 2M

wave packet

dk

d g =

== fk

p


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Q 1C Obtain an expression for the radius of mTH order bright ring in the case ofNewtons rings. 5 M

Ans.1C Figure with explanation and all the quantities shown ------------------------1 M

Write / show the interfering beams anddeduce the path difference between the

interfering beams. ------------------------1M

For Constructive interference in this case

2d = (m - ) (n = 1 for air film) ------ M

------------------------ 2M

Substituting d in the equation 2d = (m - ) , we get

---------------- M

Q2A Explain Rayleighs criterion for resolving images due to a circular aperture. 2M

Ans. 2A Raleighs criterion for optical resolution: The images of two closely spaced

sources is said to be just resolved if the angular separation of the two point sources is such

that the central maximum of the diffraction pattern of one source falls on the first minimum

of the diffraction pattern of the other. ------------------------1M

----- M

R is the smallest angular separation for which we can resolve the images of two objects.

------------- M

Q2B Obtain an expression for the quantized rotational energy of a diatomic molecule. 3 M

Ans. 2B A diatomic molecule aligned along an x-axis as shown in figure has only tworotational degrees of freedom, corresponding to rotations about the y- and z-axes.

2

12

22

R

r1RRrRRd

==

R2

r

...R

r

2

1

1RRd1Rr

22

+

=


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Rotational kinetic energy of the moleculeabout these axis = EROT , where

--------------------- M

where I is the moment of inertia of the

molecule.

------------- M

Here, m1, m2 = masses of the atoms

r = inter-nuclear separation, and = reduced mass of the diatomic molecule.

Magnitude of the classical angular momentum, L = I --------------------- M

And the quantized angular momentum of the molecule, ------------ MWhere J = 0, 1, 2, . . . are the rotational quantum number.

------------1 M

Q 2C Sketch the lowest three energy states, wave-functions, probability densities for the

particle in a one-dimensional infinite potential well. 3 M

Ans. 2C

Energy states Wave functions Probability densities

Figure 1 M each---------------(13) = 3 M

Q2D Explain (i) spontaneous emission (ii) stimulated emission with reference to lasers. 2M

Ans. 2D Spontaneous emission: The atoms excited to higher energy states are unstable

there. The excited atoms from these states spontaneously make transition to lower energystates emitting a photon whose energy h is the difference between the two energy statesE2 and E1 i.e, E2 - E1. This type of transition of an atom from a higher to a lower energy

2

21

ROTIE =

22

21

21 rrmm

mmI =

+

=

21

21

mm

mm

+=

h)1J(JL +=

( )2221

ROTI

I2

1IE.,e.i ==

( )I2

)1J(J

I2

L22

h+== )1J(J

I2

2

+=h

n


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state without any outside stimulus is called spontaneous emission. The photons so emittedare in random phases and in random directions. -----------------Figure + Definition 1 M

Excited atoms E2

emitted photon Incident photon

emitted photons

E1 spontaneous emission stimulated emission

Stimulated emission: When a photon of energy h = E2E1 is incident on an atom which isalready in an excited state, say E2, the atom being disturbed or stimulated by the incidentphoton, makes a transition to a lower energy state E1 emitting a photon. The emittedphoton has the same frequency, phase & direction as the incident photon. This type of

emission is called stimulated emission. -----------------Figure + Definition 1 M

Q3A Give the physical interpretation of spin magnetic quantum number, ms. 3 M

Ans. 3A The experimental evidence suggested the electron to have some intrinsic angular

momentum. To quantitatively indicate this quantized energy, spin quantum number is

introduced. This originates from Diracs relativistic treatment for electron. ------------ M

There can be only two directions for the

spin angular momentum vector spin-upand spin-down as shown in the figure.

Spin is an intrinsic property of a

particle, like mass and charge. The

spin angular momentum magnitude S

for the electron is expressed in terms

of a single quantum number (spin

quantum number), s = .

-----------------Figure + Explanation 1 M

--------------------- M

S is quantized in space as shown in the figure.

It can have two orientations relative to a z-axis,

specified by the spin magnetic quantum number

ms = .

The z-component of is

SZ = ms = /2

-----------------Figure + Explanation 1 M

( ) hh 23

1ss =+=S


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3B Assuming the expression for the allowed states of a particle in a 3- dimensional box,

derive the density-of-states function. 5 M

Ans. 3B The quantized energy of a particle (mass m) in a one dimensional box of(length L) are

Where n = 1, 2, 3, . . .

For a free-electron (mass m) in a metal cube of side L (three-dimensional box), thequantized energies are

nx, ny, nz = quantum numbers.

-------------- MEach allowed energy value is characterized by this set of quantum numbers (nx, ny,

nz ) and the spin quantum number ms (two values). Because of the macroscopicsize L of the box, the energy levels for the electrons are very close together. As aresult, the quantum numbers can be treated as continuous variables. In a three-dimensional quantum number space with axis representing nx, ny, nz, the allowedenergy states can be represented as dots located at positive integral values of thethree quantum numbers. --------------------------- M

Energy equation in 3D-box can be written as

-------------------- M

This is the equation of a sphere of radius n.

Thus, the number of allowed energy states

having energies between E and E+dE is equal

to the number of points in a spherical shell of

radius n and thickness dn. The volume of this

shell (NUMBER OF STATES FROM E TO

E+dE) is --------------------------- M

G(E) dE = ()(4n2dn) = ()n2dn ------------------ M

Since all the nx, ny, nz can have positive values only in an octant of the three-

dimensional space.

The number of states per unit volume [v in normal space] per unit energy range is

called density of states g (E)

------------------- M

Replacing n by ; we get ------------- M

2

2

22

2

2

2

n nLm2

nLm8

hE == h

( )2z

2

y

2

x2

22

nnnn

Lm2E ++

=

h

2

22

o

2

o

2

z

2

y

2

x

Lm2

Ewheren

E

Ennn

===++

h

oE

Enand =

V

)E(G)E(g =

oE

E

=

21

oo

21

E

Ed

E

EdE)E(G


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------------- M

---------------- M

To consider the spin states, each particle-in-a-box state should be multiplied by 2.

------------------ M

Where

Q 3C Explain Meissner effectin superconductors. 2 M

Ans. 3C Meissner effect is the exclusion of magnetic flux from the interior of

superconductors when their temperature is decreased below the critical temperature.------------------ M

----------- M

From Meissner effect, one can write, B = 0

or B = 0 inside a superconductor, below

TC. the superconductor is a perfect

diamagnet. ----------- M

A superconductor expels magnetic flux

by forming surface currents.

These surface currents induced in the superconductor produce a magnetic field that exactly

cancels the externally applied magnetic field inside the superconductor. ----------- M

Q 4A Monochromatic light with wavelength 538 nm falls on a slit with width 25.2 m. Thedistance from the slit to a screen is 3.48 m. Consider a point on the screen 1.13 cm

from the central maximum. Calculate the ratio of the intensity at this point to the

intensity at the central maximum. 3 M

Ans. 4A Given = 538 nm; a = 25.2 m ; = ? D = 3.48 m

tan () = 1.13 X 10-2 / 3.48 = 3.25 X 10-3 rad.

= 0.186. ------------------ M

dEEEdEEEE

EdE)E(G 2

12

3

o412

1

212

1

o

o

21

=

=

dEEmL2

dE)E(G 21

23

2

22

41

=

h

VL,dEELm

2

2dE)E(G 32

1

32

323

=

=h

dEEm

2

2dE

V

)E(GdE)E(g 2

1

32

23

h==

dEEh

m28dE)E(g 21

3

2

3

=

g(E) is called the density-of-states function.


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= (2 / )(a sin ) /2 = 0.478 rad = 27.4 & sin = 0.460. ------------------1 M

= msin2() / 2 . = msin

2(27.4) / 0.4782 = m0.926. ------------------1 M

i.e., / m = 0.926

Q 4B How many complete fringes appear between the first minima of the fringe envelope oneither side of the central maximum for a double-slit pattern if = 557 nm, d = 0.150 mm, and a =0.030 mm? 2 M

Ans 4B Given = 557 nm, d = 0.150 mm, and a = 0.030 mm

We have for diffraction minima,

a sin = p where p = 1, 2, 3 ------ ------------------ M

and for interference maxima,

d sin = m where m = 0, 1, 2, 3 ------ ------------------ MSince d / a = 5 , above equations suggest 5th order interference maxima (m = 5) coinciding

with the first diffraction minima (p = 1). 5th order interference fringe will be missing. 9

complete fringes will appear, 4 each on both sides of the central maximum (corresponding

to m = 1, 2, 3, 4) +one at the center corresponding to m = 0. Total 9 complete fringes.

------------------1M

4C Electrons bombard a molybdenum target, producing both continuous andcharacteristic x-rays. If the accelerating potential applied to the x-ray tube is

50.0 kV, what values of (i) MIN (ii) K (iii) K result ? The energies of the

K-shell, L-shell and M-shell in the molybdenum atom are 20.0 keV, 2.6 keV,

and 0.4 keV respectively. 3 M

Ans. 4C Given V = 50.0 10 3 V ; c = 3.00 10 8 m/s ; e = 1.60 1019 C ;h = 6.63 1034 J.s

(i)

------------------( + ) = 1 M

(ii) ------------------ M

------------------ M

.pm25m1049.21050106.1

1031063.6

Ve

ch 11319

834

MIN ==

==

nfni EEch

fh =

=

.pm63m103.6106.110)4.00.20(

1031063.6

EE

ch 11

193

834

MKk

==

=

=


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(iii)

------------------( + ) = 1 M

Q 4D When red light in vacuum is incident at the polarizing angle on a certain glass slab,

the angle of refraction is 31.8. What are (i) the index of refraction of the glass and (ii)the polarizing angle? 2 M

Ans. 4D Given r= 31.8

(ii) We have p + r = 90 ------------------ M

p = 58.2 ------------------ M

(i) n = tan p ------------------ M

and n = tan 58.2 = 1.6 ------------------ M

Q 5A A grating has 40000 ruling spread over76 mm. What is its expected dispersion D in

/nm for sodium light of wavelength 589 nm in the first order? 3 M

Ans. 5A Given order m = 1 ; N = 40000 ; Width, W = 76 103 m ; = 589 109 m

Distance between the successive slits in grating, d = W / N = 1.9 106 m ------------- M

d sin = m ------------------ M

and for first order = sin1 ( / d) = sin1 (0.31) = 18.1 ------------------ M

------------------ M

Solving, we get Dispersion D = 5.54 rad / nm ------------------ M

= 0.032 / nm ------------------ M

Q 5B An alpha particle in a nucleus can be modeled as a particle moving in a box of

length 1.0 x 1014 m. Using this model, estimate the wavelength and momentum of the

alpha particle in its first excited energy state (n=2). 2 M

Ans. 5B Given L = 1.0 x 1014 m ; n=2

For a particle in infinite potential well (box), L = n / 2 ------------------ M

Wavelength of the alpha particle in its first excited energy state, = L = 1.0 x 1014 m

---------------------- M

Momentum of a quantum particle, p = h / ------------------ M

.pm71m101.7106.110)6.20.20(

1031063.6

EE

ch 11193

834

LK

k ==

=

=

c

o

s

c

o

s

c

o

s

c

o

s

dddd

mmmm

==D,Dispersion


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11/14

Momentum of the alpha particle in its first excited energy state, p = 6.63 x 1020 kg-m/s

---------------------- M

Q 5CThe frequency of photon that causes v = 0 to v = 1 transition in the CO

molecule is 6.42 x 10

13

Hz. Ignore any changes in the rotational energy. (i)Calculate the force constant, k for this molecule. (ii) What is the maximum classical

amplitude of vibration for this molecule in the v = 0 vibrational state? 3 M

Ans. 5C Given f = 6.42 x 1013 Hz ; Mass of Carbon = 12 u; Mass of Oxygen = 16 u

1 atomic mass unit, u = 1.66 X 1027 kg ; vibrational quantum numbers, v = 0 to v = 1

Reduced mass of CO molecule is .

---------------------------- M

(i) We have

(ii) The maximum elastic potential energy = kA2 ---------------------- M

For v = 0 vibrational state, Ev=0 = hf ---------------------- M

---------------------- MSolving we get,

---------------------------- M

---------- M


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Or

Write vibrational frequency = frequency of the photon absorbed or emitted in vibrational

spectra.

i.e. kA2 = hf and solving forA, by directly substituting forf, we get

A = 4.79 pm. ---------------------1 M

Q 5DThe lifetime of an electron in the state n = 2 in hydrogen is 10 ns. What is the

uncertainty in electron energy in this state? 1 M

Ans 5D Given t = 10 x 109 s and n =2 (this data not needed)

We have E t = h / 4 ----------------------1 M

Solving forE, we get E = 33 neV ---------------------- M

Q 6A Calculate the probability of finding the electron in the ground state of an infinite

potential well of length L, between x = 0 and x = L /4. 3 M

Ans 6A The probability density is given by ( )2

nn x)x(P = ---------------------- M

i.e.,

2

nL

xnsin

L

2)x(P

=

Given n = 1 ---------------------- M

Probability for finding the electron in the x = 0 to x = L / 4 is

---------------------- M


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=

=

4/Lx

n

0x

dx)x(P ====

====

====

L/4x

0x

dxL

x(1)sin

L

22 ---------------------- M

====

====

====

4/Lx

0x

dx2

L

x2cos1

L

2

----------------------M

4/Lx

0x

4/Lx

0x

L

2

L

x2sin

L

1

L

x

====

====

====

====

====

---------------------- M

= 0.25 0.159 = 0.091 = 9.1 % ---------------------- M

Q 6B A hydrogen atom is in the first excited state (n = 2). Using the Bohr theory of

the atom, calculate (i) the radius of the orbit (ii) the kinetic energy of the

electron (iii) the potential energy of the system and (iv) the total energy of the

system. 3 M

Ans. 6B (i) We have rn = n2 a0 ---------------- M

Given n = 2 , a0

= 52.9 X 10 -12m

r2 = 2.12 X 10-10 m = 0.21 nm = 212 pm ---------------- M

(iv)En = 13.6 / n2 eV ---------------- M

E2 = 3.4 eV = 5.44 X 10-19 J ---------------- M

(iii) Potential energy = 2 X Total Energy = 6.8 eV

& (ii) Kinetic energy = 3.4 eV --------------- 1 M

(Marks should be given if students solve this numerical from basic equations)

Q 6C Most solar radiation has a wavelength of 1 m. What energy gap should thematerial in solar cell have in order to absorb this radiation ? 2 M

Ans. 6C

-------------------1 M


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---------------- M

Q 6D Calculate the energy of a conduction electron in silver at 800 K, assumingthe probability of finding an electron in that state is 0.950. The Fermi energy is5.48 eV at this temperature. 2 M

Ans. 6DGiven f ( E) = 0.950 ; T = 800 K ; EF = 5.48 eV.; Boltzmann constant, k = 1.38 X 10

23 J/ K

We have

------------------------1 M

Substituting all the values and solve forE, we get

E = 5.28 eV ------------------------1 M

( )

1kT

EE

exp

1Ef

F

+

=

i be qp makeup exam 15-07-2011 ns scheme

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