Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

JEE : Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

The document Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q.186. Calculate the Debye temperature for iron in which the propagation velocities of longitudinal and transverse vibrations are equal to 5.85 and 3.23 km/s respectively.

Ans. We proceed as in the previous example. The total number of modes must be 3nv (total transverse and one longitudinal per atom). On the other hand the number of transverse modes per unit frequency interval is given by

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

.while the number of longitudinal modes per unit frequency tnterval is given by

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

The total number per unit frequency interval is

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

If the high frequency cut off is at Irodov Solutions: Molecules and Crystals- 2 Notes | EduRevthe total number of modes will be

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Here n0 is the number of iron atoms per unit volume. Thus

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

For iron

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

(p = density, M = atomic weight of iron NA = Avogadro number).

no = 8.389 x 1022 per cc

Substituting the data we get

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 

Q.187. Evaluate the propagation velocity of acoustic vibrations in aluminium whose Debye temperature is Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev = 396 K.

Ans. We apply the same formula but assume  Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev  Then

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

or Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

For Al

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Thus                v = 3.39 km/s .

The tabulated values are  Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

and Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 

Q.188. Derive the formula expressing molar heat capacity of a unidimensional crystal, a chain of identical atoms, as a function of temperature T if the Debye temperature of the chain is equal to Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev. Simplify the obtained expression for the case T >> Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev.

Ans. In the Debye approximation the number of modes per unit frequency interval is given by

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

But  Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Thus Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

The energy per mode is Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Then the total interval energy of the chain is

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

We put ln0 it = for 1 mole of the chain.

Then Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Hence the molar heat capacity is by differentiation

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

when  Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 

Q.189. In a chain of identical atoms the vibration frequency ω depends on wave number k as ω = ωmax  sin (ka/2), where ωmax is the maximum vibration frequency, k = 2π/λ is the wave number corresponding to frequency ω, a is the distance between neighbouring atoms. Making use of this dispersion relation, find the dependence of the number of longitudinal vibrations per unit frequency interval on ω, i.e. dN/dω, if the length of the chain is l Having obtained dN/dω, find the total number N of possible longitudinal vibrations of the chain. 

Ans. If the chain has N atoms, we can assume atom number 0 and N +1 held ficed. Then the displacement of the nth atom has the form

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Here Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev Allowed frequencies then have the form 

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

In our form only +ve k values are allowed.

The number of modes in a wave number range dk is

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 But

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Hence Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

So  Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

(b) The total number of modes is

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

i.e. the number of atoms in the chain.

 

Q.190. Calculate the zero-point energy per one gram of copper whose Debye temperature is Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev = 330 K. 

Ans. Molar zero point energy is Irodov Solutions: Molecules and Crystals- 2 Notes | EduRevThe zero point energy per gm of copper is Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev is the atomic weight of the copper.
Substitution gives 48-6 J / g m .

 

Q.191. Fig. 6.10 shows heat capacity of a crystal vs temperature in terms of the Debye theory. Here Ccl is classical heat capacity, Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev is the Debye temperature. Using this plot, find: 

(a) the Debye temperature for silver if at a temperature T = 65 K its molar heat capacity is equal to 15 J/(mol•K);
 (b) the molar heat capacity of aluminium at T = 80 K if at T = 250 K it is equal to 22.4 J/(mol•K);
 (c) the maximum vibration frequency for copper whose heat capacity at T = 125 K differs from the classical value by 25%

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Ans. (a) By Dulong and Petit's law, the classical heat capacity is 3 R - 24-94 J/K - mole. Thus

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

From the graph we see that this

value of Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev corresponds to Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Hence Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 (b) 22.4 J/mole - K corresponds toIrodov Solutions: Molecules and Crystals- 2 Notes | EduRev From the graph this corresponds to

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev This givesIrodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Then 80 K corresponds to Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

The corresponding value of Irodov Solutions: Molecules and Crystals- 2 Notes | EduRevis 0.42. Hence C = 10.5 J/mole - K .

(c) We calculate © from the datum tha Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev = 0.75 at T = 125 K .

The x-coordinate corresponding to 0.75 is 0.40. Hence

 Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 Now      Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

So  Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 

Q.192. Demonstrate that molar heat capacity of a crystal at a temperature T << Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev, where Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev is the Debye temperature, is defined by Eq. (6.4f). 

Ans. We use the formula (6.4d)

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

In the limit Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev the third term in the bracket is exponentially small together wil derivatives.
Then we can drop the last term

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 Thus

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRevIrodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Now from the table in the book

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Thus Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Note Call the 3rd term in the bracket above - U3 Then

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

The maximum value of  Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev is a finite + v e quantity C0 for 0 ≤ x < Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev Thus

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

we see that Uis exponentially small as  Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 

Q.193. Can one consider the temperatures 20 and 30 K as low for a crystal whose heat capacities at these temperatures are equal to 0.226 and 0.760 J/(mol- K)? 

Ans. At low temperatures Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev. This is also a test of the “ lowness ” of the temperature We see that

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Thus T3 law is obeyed and T1, Tcan we regarded low .

 

Q.194. Calculate the mean zero-point energy per one oscillator of a crystal in terms of the Debye theory if the Debye temperature of the crystal is equal to Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Ans. The total zero point energy of 1 mole of the solid is Irodov Solutions: Molecules and Crystals- 2 Notes | EduRevDividing this by the number of modes 3N we get the average zero point energy per mode. It is Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 

Q.195. Draw the vibration energy of a crystal as a function of frequency (neglecting the zero-point vibrations). Consider two cases: T = Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev/2 and T = Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev/4, where Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev is the Debye temperature.

Ans. In the Debye model

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 Then      Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev (Total no of modes is 3N) 

Thus  Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

wc get  Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev  ignoring zero point energy

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Thus  Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev for Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 For

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRevPlotting then we get the figures given in the answer.

 

Q.196. Evaluate the maximum values of energy and momentum of a phonon (acoustie quantum) in copper whose Debye temperature is equal to 330 K.

Ans. The maximum energy of the phonon is

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

On substituting Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev - 330 K .

To get the corresponding value of the maximum momentum we must know the dispersion relation Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev . For small Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev we know Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev where v is velocity o f sound in the crystal. For an order of magnitude estimate we continue to use this result for high Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev Then we estimate v from the values of the modulus of elasticity and density

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

We write  Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Then Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Hence  Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 

Q.197. Employing Eq. (6.4g), find at T = 0:
 (a) the maximum kinetic energy of free electrons in a metal if their concentration is equal to n;
 (b) the mean kinetic energy of free electrons if their maximum kinetic energy Tmax  is known. 

Ans. (a) From the formula

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

the maximum value Emax of E is determined in terms of n by

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

or           Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

(b) Mean K.E. < E >is

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev 

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRevIrodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 

Q.198. What fraction (in per cent) of free electrons in a metal at T = 0 has a kinetic energy exceeding half the maximum energy? 

Ans. The fraction is

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 

Q.199. Find the number of free electrons per one sodium atom at T = 0 if the Fermi level is equal to EF = 3.07 eV and the density of sodium is 0.97 g/cm3

Ans. We calculate the concentration n of electron in the Na metal from

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

we get from

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRevIrodov Solutions: Molecules and Crystals- 2 Notes | EduRev

From this we get the number of electrons per one Na atom as

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

where p = density of Na, M = molar weight in gm of Na, NA = Avogadro number 

we get

0.963 elecrons per one Na atom.

 

Q.200. Up to what temperature has one to heat classical electronic gas to make the mean energy of its electrons equal to that of free electrons in copper at T = 0? Only one free electron is supposed to correspond to each copper atom. 

Ans. The mean K.E. of electrons in a Fermi gas is Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev. This must equal Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev Thus

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

 We calculate Efirst For Cu

Irodov Solutions: Molecules and Crystals- 2 Notes | EduRev

Then EF = 7.01 eV 

and T = 3 -2 5 x 104 K

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