April 09 Paper & Solution

8/12/2019 April 09 Paper & Solution

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SID ___________________ (Do not write anything else on the Q. Paper)

SHOW ALL STEPS OF CALCULATION & WRITE APPROPRIATE UNITS AT EACH STEP

PY 201 PHYSICSII Mid Term examination April 2009

Attempt all questions Time allowed: 1 hour Max. marks: 30

1. A piece of wood from an Egyptian Tomb of the old kingdom contains 1g of carbon whose

measured activity is 3.9 x 1012

Ci. Samples of fresh carbon from trees have an abundance

of 1.3 x 1010 % of

14C having a half-life of 5730 years. How old is the wood?

6

2. Energy released per proton in fusion is 6.6 MeV. The Sun contains about 1.5 x 1030

kg of

hydrogen and it radiates heat and light at the rate of 3.9 x 1026

W. How long will it take for

the Sun to burn whole of its hydrogen?

4

3. What fraction of molecules, in an ideal gas in equilibrium, has the speed within 1% of the

most probable speed vp? Find the best possible value without getting involved into any

integration. 6

4. At which temperature will the average molecular kinetic energy in gaseous hydrogen equal

the binding energy of a hydrogen atom? 4

5. A system consists of three identical particles with a total energy of 4. The energies of

states available to each particle are given by n (n = 0, 1, 2, 3, .). Enumerate

diagrammatically the possible distributions of the three particles among various energy

states assuming the particles to be (a) Fermions, (b) Bosons, and (c) Classical Particles. 10

For Honours Students Only (Additional Time Allowed : 15 minutes)

(To be attempted on a separate sheet provided after the expiry of normal allowed time)

H. (a) Is it possible for another nucleus to fuse with a56

Fe nucleus? Give reasons. How did

nuclei with A > 56 originate?

(b) What is Dulong-Petit law? What are its failures?

(c) How did Einstein approach the problem of specific heat of solids? What is his formula

for the specific heat? How does it stand upto the experimental results?

3

3

4


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Solution Hints PY 201 PHYSICSII Mid Term examination April 2009

1. Total no. of atoms in 1g of carbon = 2223

1002.512

10023.6

atoms

No. of14

C atoms in 1g of a sample of fresh carbon,N0 =1022

10

1053.61002.5100

103.1

atoms

Decay constant of14C, 12

2/1

10835.3sec3600243655730

693.05730

693.0693.0

yrT

s1

Activity of 1g of fresh carbon sample due to14

C, 25.01053.610835.3 101200

NA s1

Activity of 1g of old wood sample, 101212 107.3109.3109.3 CiA s1 21043.14 s1

We know that 1110835.320

10433.125.01043.14,12

teeAA tt

s = 4544 yr

2. Total no. of atoms in 30105.1 kg of hydrogen = 563026

1003.9105.1

1

1002.6

atoms

OR the number of protons in the sun = 561003.9 protons

Energy released per proton in fusion = 6.6 MeV = 13106.16.6 J 1210056.1 J

Total energy content of the sun in regard to fusion of protons = 561003.9 4412 1054.910056.1 J

Rate at which energy is radiated by the sun = 3.9 x 1026 W = 3.9 x 1026 J/s

Time it will take the sun to burn whole of its hydrogen = 1826

44

1045.2/109.3

1054.9

sJ

Js = 7.77 1010 yr

3. The most probable speed of a molecule is given bym

Tkvp 2

The M-B law of distribution of molecular speeds is

dvevkT

mNdvvn kTmv 2/2

23

2

24)(

(N= Total no. of molecules)

the fraction of molecules having speeds in the range v to v + dv is given by

dvevkT

m

N

dvvndvvf kTmv 2/2

23

2

24

)()(

We wish to find the fraction of molecules having speeds within 1% ofvp. The exact way to do this

would be to integrate the above expression within appropriate limits. However, an approximate value canbe found by calculatingf(v) with v = vp and multiplying it with the total spread of speeds.

1

2/3

2

24)(

e

m

Tk

Tk

mvfp

Spread of speedsm

Tk

m

Tkvvv pp

2

50

12

100

2

100

11

100

11

the required fraction of molecules =m

Tke

m

Tk

Tk

mvvf p

2

50

12

24)(

1

2/3

=e50

4 = 0.0166


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4. Binding energy of hydrogen atom = 13.6 eV = 19106.16.13 J = 1810176.2 J

Average molecular kinetic energy = Tk2

3

Equating the two, Tk2

3= 1810176.2

KJ

JT

/1038.13

10176.2223

18

= 51005.1 K

5. The individual particle states have energies 0, , 2, 3, 4, .

Since the total energy of the system is 4, none of the states with energy > 4 can be populated. So we

restrict ourselves to the first five states upto energy 4 only.

For total energy of the particles to be 4, there are only four possible ways they can be distributed among

the various states:

(i) one particle in each of the states of energy 0, and 3

(ii) one particle in state of energy 0 and the remaining two particles in the state of energy 2

(iii) Two particles in the 0-energy state and the third one in state of energy 4

(iv) Two particles in state of energy and the third in state of energy 2

(a) When the particles are fermions, they are indistinguishable and only one particle can occupy one

state (Pauli exclusion principle). If we represent each particle by X, there is only one way the three

particles can be in separate states and have a total energy of 4. This is sketched below:

(b) Bosons are again indistinguishable but there is no restriction on the number that may occupy any

state. Therefore, there are four different distributions possible as shown below:

(c) Classical particles, like bosons, do not obey the Pauli exclusion principle but they are

distinguishable. Therefore, in addition to the number of particles in a state, it is imperative to specify

which particle is in which state, to uniquely define a distribution. Let us represent the three particles

as X1, X2 and X3. There are, now, fifteen different distributions having the total energy 4. These canbe derived from the four distributions of bosons as given above by interchanging the particles among

themselves (keeping the number of particles in a state fixed):

Energy

stateDistribution of particles

0 X X2 --3 X4 --

Energy

state

Distributions of particles

1 2 3 4

0 X, X X X -- -- X -- X, X2 -- -- X, X X3 -- X -- --4 X -- -- --


4/4

Energy

state

Distributions of particles

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0X1X2

X1X3

X2X3

X1 X1 X2 X2 X3 X3 X1 X2 X3 -- -- --

-- -- -- X2 X3 X1 X3 X1 X2 -- -- -- X1X2

X1

X3

X2

X3

2 -- -- -- -- -- -- -- -- -- X2X3

X1X3

X1X2

X3 X2 X1

3 -- -- -- X3 X2 X3 X1 X2 X1 -- -- -- -- -- --4 X3 X2 X1 -- -- -- -- -- -- -- -- -- -- -- --

Question for Honours students

H. (a)56

Fe is the nucleus with the greatest B.E. per nucleon. Fusion of another nucleus with this will result

in a nucleus having lesser BE per nucleon, and hence is not energetically possible.

Nuclei with A > 56 are believed to have originated through successive capture of neutrons, followed

by beta decays to achieve appropriate neutron/proton ratios.

(b) The specific heat of a solid may be estimated classically assuming each atom (ion/molecule) to

behave as a 3-dimensional oscillator with average energy kT per dimension (independent of frequency

of the oscillator). This approach gives the molar specific heat of a solid at constant volume to be

Cv = 3 R = 5.97 kcal/kmol-K

This is known as Dulong-Petit law as Dulong and Petit had discovered experimentally that Cv 3R

for most solids at normal temperatures.

o Dulong-Petit law fails for all solids at very low temperatures. It is found that Cv 0 for all solids

as T 0 K.

o It fails for light elements like boron, beryllium and diamond, even at normal temperatures.

(c) Einstein considered the classical law of equipartition of energy to be flawed. According to him, the

probability f() for an oscillator to have frequency is given by f() = 1 / (eh/kT1). The average

energy of an oscillator then comes out to be h/ (eh/kT 1), and not kT. Accordingly, Einsteins

formula for specific heat of a solid is

2

2

1

3

kTh

kTh

v

e

e

kT

hRC

It can be seen that it reduces to Dulong-Petit law at high temperatures.

However, the variation in Cv as T 0, predicted by this formula, does not match the exact manner of

variation observed experimentally.

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April 09 Paper & Solution