Final Mp 2013 Soln

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Total marks of part A: 71 Total Time:3 hours

Final Exam 2013: Modern Physics

Solution

Name: Roll no:

Write your name and roll number in the space specified above.

This exam comprises two parts, A and B. Part A comprises 23 questions. The mostappropriate answer is to be circled on the question paper. Part A is to be filled on the

question paper and returned.

For answering part B, you will use the provided blue answer books. There are twoquestions in part B.

Part A contains 15 pages including this page.

Part B contains 2 pages.

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Fundamental constants and other useful information

c = Speed of light = 3.0 108 ms1

h = Plancks constant = 6.63 1034 Js = Reduced Plancks constant =

h

2= 1.06 1034 Js

kB = Boltzmann constant = 1.38 1023 J/KR = Rydberg constant = 1.1 107 m1

me = Mass of electron = 9.11 1031 kgmP = Mass of proton = 1.67 1027 kg

1eV = 1.6 1019 Jg = Acceleration due to gravity = 10.0 m s2

G = Gravitational constant = 6.67 1011 N m2/kg21

40= Coulombs constant = 9 109 N m2/C2

TISE : 2

2m

d2(x)

dx2+ V(x)(x) = E(x)

TDSE : 22m

d2(x, t)

dx2+ V(x)(x, t) = i

d

dt(x, t)

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PART A

Attempt all questions. Mark your answers on these sheets and return. All MCQs are three

marks each.

1. A metal is held at zero voltage. The energy diagram at the metal-air interface is

shown.

EF

Metal Air

Unfilled levels

Filled levels

In thermionic emission, electrons are ejected from the metal surface because:

(a) The work function increases.

(b) The work function decreases.

(c) The potential energy seen by the electrons in the air slopes downward.

(d) Increasing temperature makes more electrons jump into unfilled levels increasing

the fraction of electrons with thermal energy beyond .

(e) The Fermi level EF decreases.

Answer 1:

In thermionic emission, heating increases the thermal energy of the electrons. These

electrons are raised from the filled to the unfilled levels. Some of these excited electrons

obtain enough energy to overcome the work function and can therefore be ejected into

air.

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2.

EF

MetalOxide

layer

Quantum dot

i

e /2Cdot2

1

2

34

56

7

89

10

n (quantum number)

The figure shows the energy diagram for a metal in which electrons fill energy levels

upto EF. A thin insulating oxide layer separates the metal from a quantum dot with

only ten quantized energy levels. The quantum dot is given a positive potential V0

with respect to the metal, enabling an electron to tunnel across the oxide layer. Which

one of these plots shows the correct behavior of the tunneling current i from metal to

the quantum dot. At V0 = 0, EF is at the same energy as the n = 7 quantum level.

Vo

i(a)

Vo

i(b)

e /2CdotVo

i(c)

e /2Cdot

Only three peaks

are observed

Vo

i(d)

e /2Cdot

Only four peaks

are observed

Vo

i(e)

e /2Cdot

Ten peaks are

observed

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Answer 2:

Option (d) is the correct answer. The quantum dot is given a variable positive potential

V0. An electron added to the quantum dot raises its coulomb energy by e2/2Cdot.

Hence if energy is to be conserved and the electron transfer to the dot is to be favored,the starting energy of the dot must be lower by e2/2Cdot, so that pickup of an extra

electron is energetically permissible. IfV0 = 0, the electron cannot tunnel into n = 7 as

it will raise the overall energy of the dot. If, however, V0 = e/2Cdot, n = 7 is lowered in

energy by e2/2Cdot and electron tunneling to n = 7 becomes energetically permissible.

The electron tunnel! While keeping its total energy constant. The quantum dot is

a receptacle lowering its energy in anticipation of an incoming electron, which raises

the energy back to the original. Hence four peaks corresponding to tunneling ton = 7, 8, 9, 10 are observed.

3. An electron is injected into a potential energy landscape from the left region I as shown

below. It encounters a potential step. The energy of the electron is E and E < |V0|.If the electron is to emerge in region III with a faster speed, the appropriate potential

step is given by which of the following?

(a)

E

Vo

(b)

E

Vo

(d)

E

Vo

(c)

EVoI

II

III I

II

III

I

II

III

I II III

(e) The speed of the electron cannot incraese.

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Answer 3:

Option (c) is the correct answer. If v is to go up in region III, k must increase. Since

k E V, the difference between E and V must be higher, which is the situation

in Fig.(c).

4.

1 mm

A pendulum is a harmonic oscillator. It completes one round trip in 1 s. According

to quantum physics, its minimum energy in Joules is,

(a) Zero. (b) 6.63 1034.(c) I need to know the angle to answer this question.

(d) 3.315 1034. (e) It is negative.Answer 4:

Minimum energy of a harmonic oscillator is,

Emin. =1

2,

where = h/2 and = 2/T, where T = 1 s is the time period of harmonic

oscillator. Therefore,

Emin. =1

2

h

2

2

T

=1

2

h

T

=1

2

6.63 1034 Js

1 s

= 3.315 1034 J.

Hence option (d) is the correct answer.

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5. Suppose the maximum angle of the pendulum discussed in the previous question is

such that the bob goes to a maximum height of 1 mm. The mass of the pendulum is 1

gram and g = 10 ms2. The bob starts oscillating and eventually comes to rest, losing

all of its energy. Energy is lost in the form of a phonon when the oscillator makes atransition from the pth quantum level to (p1)th quantum level. How many phononsare emitted in the process?

(a) Negligible. (b) One.

(c) 1.5 1028. (d) 1.5 1033.(e) None of the above.

Answer 5:

We are given that,

Height of bob = y = 1 mm = 103 m

Mass of the pendulum = m = 1 g = 103 kg

Acceleration due to gravity = g = 10 m/s2.

the maximum energy of the photon is,

mgy =

n + 12

=

n +

1

2

h

2

2

T

=

n +

1

2

h

T,

and the principle quantum number is,

n =

mgy

h 1

2

=103 kg 10 m/s2 103 m

6.63 1034 Js 1

2

1.5 1028.

As the pendulum loses all of its energy, the quantum number decrements by 1, i.e.,

from n to n 1 to n 2 to n 3 all the way to n = 0. On each decrement, phonon isemitted. Hence the total number of phonons emitted is 1.51028. Therefore option(c) is the correct answer.

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6. Snells law of refraction determines the bending of light across an interface. For sure,

electrons are also waves and can be refracted. The corresponding law for electrons is

called Bethes law and is given by,

sin sin

= v2v1

,

where is the angle of incidence measured from the normal to the interface, is the

angle of refraction also measured from the normal, v1 is the speed of electron in the

incident medium and v2 is the speed in the refracted medium. Now a beam of electrons

is made to pass through two hollow cylinders with an applied voltage difference. Which

of the following diagrams show the correct trajectory of electrons?

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++++++

++++++

Incidentmedium

refracted

medium

Electron gun

(a) No bending.

++++++

++++++

Incidentmedium

refracted

medium

Electron gun

(b) Bending towards normal.

++++++

++++++

Incident

medium

refracted

medium

Electron gun

(c) Bending away from normal.

++++++

++++++

Incident

medium

refracted

medium

Electron gun

(d) No refraction takes place.

(e) None of these.

Answer 6:

In the refracting medium, the electric potential is positive and hence the potential

energy seen by the electron, V, is lower. This means that the difference E V islarger, k is larger, and hence the speed is slower, v2 < v1. Hence from Bethes law,

sin < sin , < . The electron beam bends away from the normal. Hence option

(c) is the correct answer.

7. A proton of rest mass m0 is accelerated to half the speed of light. Its de Broglie

wavelength is,

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(a)

3h/(m0c). (b) 2h/(

3m0c).

(c) 4h/(

3m0c). (d) 4h/(m0c).

(e) A proton does not have a wavelength.

Answer 7:

From the de Broglie hypothesis we know that wavelength associated with the matter

wave is,

=h

p

=h

mv, since p = mv,

where according to special relativity if m0 is the rest mass of the proton,

m =m01 v2c2

= hm0v

1 v

2

c2.

Since we are given that, speed of proton v = c/2,

= 2hm0c

1 c

2

4c2

=2h

m0c

3

2

=

3h

m0c.

Hence option (a) is the correct answer.

8. An electron starts off in the region B, trapped in a well. The potential energy V(x)

along position x is shown.

x

Region CRegion A Region B

E= Energy of the electron

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Now suppose some time lapses. From a quantum viewpoint, which of the regions A or

C, is the electron more probable to be found?

(a) Region A.

(b) Region C.

(c) Equal probability of being found in A and C.

(d) The electron absolutely cannot leave region B.

(e) None of the above.

Answer 8:

As we know that when a quantum object encounters a wider barrier, the tunneling

transmission probability is lower. If the barrier is thinner the tunneling probability is

higher. In the given figure we can see that the barrier on the right is thinner. Therefore

it is more probable to find the electron in region C as compared to A. Hence option

(b) is the correct answer.

9. A particle moving in a region of zero force encounters a precipicea sudden drop in

the potential energy to an extremely large negative value. What is the probability

that it will go over the edge, i.e., it will enter the negative potential energy region?

(a) Almost zero. (b) Almost one.

(c) 1/2. (d) > 1/2.


Answer 9:

One need to think carefully about this. Consider the accompanying figure.

Vo

E

The potential depression V0 is large. Lets find the reflection probability R. In region

I,

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I(x) = Aeik1x + Beik1x,

and for region II,

II(x) = Ceik2x,

where k21 =2mE

, and k2 =

2m(E+V0)

. At the point of the precipice, x = 0, I(0) =

II(0) and I(0) =

II(0). So A + B = C and ik1(A B) = ik2C. Eliminating C

from these equations,

A + B =ik1(A B)

ik2

=k1k2

A k1k2

B

B

1 +

k1k2

= A

k1k2 1

B

A=

k1 k2k1 + k2

R =|B|2|A|2

k1k1

=k1 k2k1 + k2

2

.

IfV0 is very large, k2 k1, R becomes(k2k2

)2= 1. Since R 1, T = 0. There is zero

probability for the particle to fall over the edge and enter region II. Hence option

(a) is the correct answer.

10. An LED (light emitting diode) emits light when electrons fall from a top set of levels

(conduction band) to a bottom set of levels (called a valence band). These levels are

separated by an energy gap Eg. An electric current supplies the electrons.

Hole flow

Electron flow

Eg= hf Photon

= Electron

= Hole

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A current of 2.5 mA flows through an LED with Eg = 1.4 eV. Assuming that each

current-carrying electron drops into a hole, thereby emitting a single photon, what is

the power emitted in the light?

(a) 5.6 1022 W. (b) 3.5 mW.(c) 2.5 mW. (d) 1.4 mW.

(e) The power cannot be calculated using the provided information.

Answer 10:

we are given that,

Current through LED = I = 2.5 mA = 2.5

103 A

Energy gap = Eg = 1.4 eV = 1.4 1.6 1019 JNo. of electrons flowing per unit time =

I

Charge of electron=

I

e

No. of photons emitted per unit time =I

Charge of proton=

I

e

Energy of each photon = Eg.

Hence power emitted in the light is,

P = Ie Eg

P =2.5 103 A1.6 1019 C 1.4 1.6 10

19 J

= 3.5 103 J/s= 3.5 mW.

Hence option (b) is the correct answer.

11. An electron is trapped inside a three-dimensional quantum dot. The energy is quan-

tized in three dimensions according to,

Enx,ny ,nz =22

2m

n2xa2

+n2yb2

+n2zc2

,

where a, b and c are the confining dimensions of the box(= dot) and nx, ny, nz are

the three quantum numbers, each one of them being a positive integer.

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a

b

c

Quantum Dot''

''

If a = b = c, the energy difference between the ground and the first excited state is,

(a) 22

2ma2. (b) 9

22

2ma2.

(c) 3 22

2ma2. (d)

22

ma2.

(e) There are more than one first excited states all with different energies. Hence

this question cannot be answered.

Answer 11:

We are given that,

Enx,ny,nz =22

2m n2xa2

+n2yb2

+n2zc2

=22

2m(n2x + n

2y + n

2z) since a = b = c.

For ground state (nx, ny, nz) = (1, 1, 1), energy of the ground state will be,

E(1, 1, 1) =22

2m(12 + 12 + 12)

= 322

2m.

Similarly for first excited state, (nx, ny, nz) = (2, 1, 1), energy of the first excited state

will be,

E(2, 1, 1) =22

2m(22 + 12 + 12)

= 622

2m.

Energy difference for these two energy levels is,

E(2, 1, 1) E(1, 1, 1) = 622

2m 3

22

2m

= 322

2m

.

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Hence option (c) is the correct answer.

12. A free particle has a wave function,

(x, t) = Ae

i(2.5

1011x

2.1

1013t)

,

where x is in metres and t is in seconds. What is the mass of the particle?

(a) Mass can only be determined if A is known.

(b) 0.012 kg. (c) 0.11 kg.

(d) 5.7 1016 kg. (e) 1.7 1015 kg.

Answer 12:

As we know that general equation of wave function is,

(x, t) = Aeikxt.

Comparison of this equation with the given wave equation of for the free particle yields,

k = 2.5 1011 m1

= 2.1 1013 s1.

Mass of the particle can be calculated by the dispersion relation,

2 =k

m

m = k2

=2.5 1011

(2.1 1013)2= 5.7 1016 kg.


You dont really need to remember the dispersion relationship. Look at the TDSE in

the absence of V:

TDSE : 2

2m

d2

dx2(x, t) = i

d

dt(x, t).

Inserting the supplied wave function into the above, the relationship 2 = k/m au-

tomatically pops out. Students are tempted to use E = and E = p2/2m. The

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former relationship does not hold for all particles, it is specific in its meaningit says

that energy of a photon E is related to the frequency of the electromagnetic wave

associated with the photon. Blindly using these relations is wrong!

13. A magic-eye was shown as a classroom demonstration. In this gadget, a beam of

electrons is emitted from a cathode and moves horizontally towards a bowl shaped

anode. When viewed from top, the electrons produced a characteristics glow as shown.

(a) (b)

AnodeBeam of electrons

Anode

Cathode

Electron glow

k

k = Cathode

represents magnetic field outof the plane of the paper.

show the direction of

moving electrons.

If the tube is placed inside a perpendicular magnetic field pointing out of the plane of

the paper, as shown in Fig (b), the view from the top will look like:

(a)

(b) (c)

(d) (e)

Electron glow is

shown shaded

Answer 13:

Lets find out the direction of the forceF on the beam of electrons. The force is given

byF = ev B. Using the right hand rule; pointing the fingers in the direction ofv , and curling these into the direction of B, the thumb points against the direction

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of force. (Against because of the negative sign, the electrons are negatively charged).

In the leftward beam, the beam bends upwards and in the rightward beam, it bends

downwards. Hence option (d) is the correct answer.

14. A beam of electrons accelerated through a potential difference V0 is directed at a single

slit of width a, then detected at a screen at a distance L beyond the slit. How far

from a point directly in the line of the beam is the first location where no electrons

are ever detected?

(a) L/(a

V). (b) Lh/(a

2meV).

(c) The electrons are detectable everywhere.

(d) L/a. (e) Lh/(2a

2meV).

Answer 14:

a/2

ZRay I

Ray II

L

Path difference between ray I and II is,

Path difference =a

2sin ,

and for destructive interference this path difference must be /2.

a

2sin =

2

a sin =

As is small, sin tan = zL

=

a

z =L

a.

Since = h2meV

, z = Lah

2meVHence option (b) is the correct answer.

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15. An electron of energy 1 eV is trapped inside an infinite well of length 30 cm. What is

the distance between two consecutive nodes of the electrons wavefunction? (A node

is a point where the wavefunction goes to zero.)

(a) There are no nodes in the electrons wavefunction.

(b) The distance between consecutive nodes is zero.

(c) 1.25 1018 m.

(d) 6 1010 m.


Answer 15:

We are given that,

Energy of electron = E = 1 eV = 1.6 1019 JLength of infinite well = L = 30 cm = 0.3 m.

The wave function is A sin

nxL

. The consecutive nodes differ in space by x. At

a node, the argument of the sine function must differ in phase by . Hence nxL

=

,

x = L/n. We just need to find n, the state of the wave function.

x

sin (nx/L) is zero here

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The quantum number n is calculated as,

En =n222

2mL2

n2 =

2EmL2

22

n =

2EmL2

22

=

2 1.6 1019 J 9.11 1031 kg (0.3 m)2

(1.06 1034 Js)22=

2.4 1017

= 4.9 108.

Thus the distance between two consecutive nodes is,

x =0.3 m

4.9 108= 6 1010.


16. For the electron in the previous question, will quantum effects be visible? Please give

a reason. [2 Marks]

Answer 16:

Since the length of infinite well is very large i.e. 30 cm, for a small amount of energy

1 eV, the number of nodes will be very large i.e. 4.9 108. Since the number of nodesis very large, the waves are squeezed close together, the de Broglie wavelength is

extremely small obscuring chances of observing the quantum wave behavior at the

classical macroscopic scale. At such a high value of n, quantum effects are not visible.

Another way of looking at this is that the wave function is such that the probability of

finding the electron becomes equal everywhere, i.e. it imparting the electron a contin-

uous quality rather than quantized. All of this ties in well with Bohrs corresponding

principle.

17. An electron is trapped in an infinite well of length L and ground state energy E1. At

t = 0, the wavefunction is,

(x, 0) =15L(1(x) + 22(x)),

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where 1(x) and 2(x) are normalized wavefunctions in the ground and first excited

states. The wavefunction at t = /E1 is given by:

0.2 0.4 0.6 0.8 1.0

-2

-1

1

0.2 0.4 0.6 0.8 1.0

-1

1

2

0.2 0.4 0.6 0.8 1.0

-1

1

2

0.2 0.4 0.6 0.8 1.0

-2

-1

1

0.2 0.4 0.6 0.8 1.0

1

2

3

4

(a)(b)

(c)

(d)

(e)

()Re ()Re

()Re()Re

()Re

x/L

x/L

x/L

x/L

x/L

Answer 17:

We are given that,

At t = 0 (x, 0) = 15L

(1(x) + 22(x)

)At any time t (x, t) =

15L

(1(x)e

iE1t/ + 22(x)eiE2t/).

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If E1 is the energy of ground state, energy of first excited state will be E2 = 4E1.

(x, t) = 15L

(1(x)e

iE1t/ + 22(x)ei4E1t/)

At t =

E1 (x, t) =

1

5L(1(x)ei + 22(x)ei4)=

15L

(1(x)(1) + 22(x)(+1)

)=

15L

(1(x) + 22(x)).Now we need to find out what 1(x) + 22(x) looks like. The construction is seenin the series of diagrams below.

0.2 0.4 0.6 0.8 1.0

-2

-1

1

2

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

.0

0.2 0.4 0.6 0.8 1.0

-1.0

-0.8

-0.6

-0.4

-0.2

0.2 0.4 0.6 0.8 1.0

-2

-1

12 (x)

1(x)

1 (x)A

-A

1 (x)+ 2 (x)2

2

Invert

Add


18. The potential energy profile in a certain region is shown.

a bx

oo

ooto

to

energy E

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A particle of energy E exists inside this region. A sketch of the possible (real part) of

the wavefunction is;

a bx

(a)

a bx

(b)

a bx

(c)

a bx

(d)


wavefunction

is zero at x > b

wavefunction is

non zero at x > b

wavefunction

is zero at x > b

wavefunction

is zero at x > b

Answer 18:

Option (d) is the correct answer. The wavefunction is zero at x b because of theinfinite potential and extends into the region x

a. Furthermore, the value of k

increases and wavelength decreases as we go from x = a to x = b.

19. Suppose a particle is in the ground state with wavefunction 1(x). Which one of the

following is the probability that the particle will be found in a narrow range between

x and x + dx. Furthermore, the value of k increases and wavelength decreases as we

go from x = a to x = b.

(a) |1(x)|2dx. (b) x|1(x)|2dx.

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(c)dxx

x|1(x)|2dx. (d)+

x|1(x)|2dx.


Answer 19:Option (a) is the correct answer.

20. At time t = 0, the state for a particle inside an infinite well is 12

(1 + 2), where

1 and 2 are ground and first excited states: with energies E1 and E2 respectively.

We first measure the position of the particle at time t = 0 and obtain the result x0.

Immediately after the position measurement, we measure the energy. What possible

result(s) can we obtain for the energy measurement?

(a) We can only measure either E1 or E2.

(b) We can obtain one of the energy values En =n222

2ma2, where n can be an arbitrary

large integer.

(c) We can only measure 12

(E1 + E2).

(d) We may measure any energy E =n=1

CnEn, where Cn are coefficients so that

n=1 |

Cn|2 = 1.


Answer 20:

The correct option is (a).

21. A free particle has a wavefunction A(eikx+ eikx) and energy E. A is a normalization

constant. Mark True of False against these statements.

(a) The probability density does not change with time.

(b) The probability density is constant in space x.

(c) The de Broglie wave associated with the particle is in fact a standing wave.

[6 Marks]

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Answer 21:

We are given the wavefunction of free particle,

(x, t) = A(eikx + eikx)eiEt

(x, t) = A (eikx + e+ikx)eiEt(x, t)(x, t) = A(eikx + e+ikx) A(eikx + eikx)eiEt eiEt

= A2(1 + e2ikx + e2ikx + 1)

= A2(2 + e2ikx + e2ikx).

Using cos x = eix+eix

2

|2(x, t)|2 = A2(2 + 2 cos(2kx))= A2(2 + 2 cos(2kx))

= 2A2(1 + cos(2kx))

= 2A2 2cos2(kx))= 4A2 cos2(kx))

(a) True since p(x) = |2(x, t)|2 is independent of time.

(b) False since p(x) depends on x and changes with x.

(c) True because the forward and backward propagating waves have equal ampli-

tudes and the probability density does not change with time.

22. The uncertainty relationship for a particle moving in a straight line is px /2.

R

S = Arc length

S

R = radius

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If the particle is moving in a circle with angular momentum L, the uncertainty rela-

tionship becomes:

(HINT: Distance becomes the arc length!)

(a) L 2 .(b) LS

2.

(c) LR 2

.

(d) L 2

.


Answer 22:

According to uncertainty principle,

px 2

.

If particle moves in a circle of radius r and angular momentum L, then

L = pr

L = pr

p = Lr

and x = r.

Using these values of p and x in uncertainty relation,

L

r r

2

L 2

.


23. In a scanning tunneling microscope (STM), the tunneling probability of electrons from

metal surface to a prob tip is proportional to exp(2L), where L is the tip-sampledistance and = 1 nm1 is the inverse of the penetration length.

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L

Tip

Metal surface Metal surface

Tip

LL -

If the tip moves closer to the surface by L = 0.1 nm, the tunneling current,

(a) remains unchanged.

(b) increase by 22 %.

(c) decrease by 22 %.

(d) increase by 10 %.

(e) decrease by 10 %.

Answer 23:

We are given that,

Tip-sample distance = = 1 nm1 = 1 109 m1

Distance covered by tip = L = 0.1 nm = 0.1

109 m.

Tunneling probability is,

Ti = e2L.

If tip moves closer to the surface by L, final tunneling probability will become,

Tf = e2(LL).

ratio of tunneling probabilities is,

TfTi

= e2(L

L)

e2L

= e2L

= e21090.1109

= e0.2

= 1.22.

Hence there is a 22% increase in the tunneling current and the correct answer is (b).

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Total marks of part B: 20 Total Time:3 hours

Final Exam 2013: Modern Physics

PART B

Attempt all questions.

1. The Heisenberg uncertainty principle applies to photons as well as to

material particles. Thus a photon confined to a small box of size x

necessarily has a large uncertainty in momentum and uncertainty in

energy. Recall that for a photon E= pc.

(a) Estimate the uncertainty in energy for a photon confined to the tiny

box of size x. [2 Marks]

Answer 1 (a):

We are given that,

E= pc.

Uncertainty in energy is,

E cp. (1)

Uncertainty in momentum p for a photon confined to the tiny box of

size x can be calculated by using the uncertainty relation,

x p

2,

where is the reduced Plancks constant.

p

2x.

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Using this value of p in equation (1) yields,

E c

(

2x

)

E c2x

(b) If E E, what is the effective mass of the photon? [2 Marks]

Answer 1 (b):

We are given that,

E E

E=c

2x

Effective mass can be calculated by using energy-mass relationship,

E= meffc2,

where meff is the effective mass of the photon.

meff =E

c2

=c

2x

1

c2

=

2cx.

(c) This mass can be extremely large, if x is tiny. If x is sufficiently

small, the large mass can create a large gravitational field, sufficiently

large to form a black hole. When this happens x is called the Planck

length, and this is when gravity and quantum mechanics become inter

mixed. For a black hole, not even light can escape.

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R

M

Launch an object

Consider a star of mass M and radius R as shown above. If an object

is to be launched from the stars surface so that it escapes the stars

gravitational pull, it needs a minimum velocity vesc

called the escapevelocity. Show that vesc =

2GM

R. [4 Marks]

Answer 1 (c):

We are given that,

Mass of the star = M

Radius of the star = R

Let mass of the object = m,

and let speed of the object to escape from the stars gravitational pull is

vesc. In order to escape from the gravitational pull of star, kinetic energy

of the object should be at least equal to the gravitational potential

energy. K.E. = P.E.

1

2mv2esc. = G

mM

R

v2esc. =2GM

R

vesc. = 2GM

R

.

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Which is the required result.

(d) Ifvesc = c, nothing can escape from this star, not even light. If we

were to replace the star of mass Mwith a photon of the mass calculated

in part (b), and confined to length x, and set R = x, calculate the

Planck length in terms ofG, and c. [3 Marks]

Answer 1 (d):

Using the relation ofvesc derived in part (c)

vesc =2GM

R ,

where vesc = c,

c =

2GM

R

c2 =2GM

R.

Setting R = x and M= meff,

c2 =2Gmeff.

x

=2G

x

2cx

=G

cx2

x2

=

G

c3

x =

G

c3.

(e) IfG = 6.67 1011 N m2/kg2, find the numerical value of Plancks

length. [2 Marks]

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Answer 1 (e):

x =

G

c3

=

6.67 1011 N m2/kg2 1.06 1034 Js

(3.0 108 ms1)3

=

2.62 1070 m2

= 1.6 1035 m.

(f) What is the diameter of a proton (about 2 fm = 2 1015 m) in

units of Plancks length? [2 Marks]

Answer 1 (f):

We are given that,

Diameter of photon = 2 fm

= 2 1015 mDiameter of photonin Plancks length

= 2 1015 m1.6 1035 m/Planck length

= 1.24 1020 Planck lengths.

2. The radioactive decay of certain heavy nuclei by emission of an alpha

particle is a result of quantum tunneling. Imagine an alpha particle

moving around inside a nucleus, such as thorium (mass number= 232).

When the alpha particles bounces against the surface of the nucleus, it

meets a barrier caused by the attractive nuclear force. The dimensions

of barrier vary a lot from one nucleus to another, but as representative

numbers you can assume that the barriers width is L 35 fm (1 fm

= 1015 m) and the average barrier height is such that V0 E 5

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MeV. Find the probability that an alpha hitting the nucleus surface will

escape. Given that the alpha hits the nuclear surface about 5 1021

times per second, what is the probability that it will escape in a day?

The tunneling probability is T = e2L where =

2m(V0 E)/and

L is the barrier length. (1 MeV= 106 eV). [5 Marks]

Answer 2:

We are given that,

Barrier length = L = 35 fm = 35 1015 m

Barrier height = (V0 E) = 5 MeV = 5 106 1.6 1019 J

= 8 1013 J

Tunneling probability = T = e2L,

where =

2m(V0 E)

and mass of the nucleus = m = 4 1.67 1027 kg = 6.68 1027 kg.

In order to calculate probability of escape T, lets first calculate .

=

2m(V0 E)

=

2 6.68 1027 kg 5 106 1.6 1019 J

1.06 1034 Js

= 9.75 1014 m1.

Tunneling probability is,

T = e2L

= e29.751014 m1351015 m

= e68.25

= 2.29 1030.

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The probability that an alpha particle hitting the nucleus surface will

escape is 2.29 1030.

If no. of hits per second = 5 1021

/s Probability of escape

in a day

= T = 2.29 1030 5 1021/s 24 60 60 s

= 9.89 104

= 9.89 102 %

Final Mp 2013 Soln

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