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 3n_{0 }v (total transverse and one longitudinal per atom). On the other hand the number of transverse modes per unit frequency interval is given by
.while the number of longitudinal modes per unit frequency tnterval is given by
The total number per unit frequency interval is
If the high frequency cut off is at the total number of modes will be
Here n_{0} is the number of iron atoms per unit volume. Thus
For iron
(p = density, M = atomic weight of iron N_{A} = Avogadro number).
n_{o} = 8.389 x 10^{22} per cc
Substituting the data we get
Q.187. Evaluate the propagation velocity of acoustic vibrations in aluminium whose Debye temperature is = 396 K.
Ans. We apply the same formula but assume Then
or
For Al
Thus v = 3.39 km/s .
The tabulated values are
and
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 . Simplify the obtained expression for the case T >> .
Ans. In the Debye approximation the number of modes per unit frequency interval is given by
But
Thus
The energy per mode is
Then the total interval energy of the chain is
We put ln_{0} it = for 1 mole of the chain.
Then
Hence the molar heat capacity is by differentiation
when
Q.189. In a chain of identical atoms the vibration frequency ω depends on wave number k as ω = ω_{max } sin (ka/2), where ω_{max} 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 n^{th} atom has the form
Here Allowed frequencies then have the form
In our form only +ve k values are allowed.
The number of modes in a wave number range dk is
But
Hence
So
(b) The total number of modes is
i.e. the number of atoms in the chain.
Q.190. Calculate the zeropoint energy per one gram of copper whose Debye temperature is = 330 K.
Ans. Molar zero point energy is The zero point energy per gm of copper is is the atomic weight of the copper.
Substitution gives 486 J / g m .
Q.191. Fig. 6.10 shows heat capacity of a crystal vs temperature in terms of the Debye theory. Here C_{cl} is classical heat capacity, 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%
Ans. (a) By Dulong and Petit's law, the classical heat capacity is 3 R  2494 J/K  mole. Thus
From the graph we see that this
value of corresponds to
Hence
(b) 22.4 J/mole  K corresponds to From the graph this corresponds to
This gives
Then 80 K corresponds to
The corresponding value of is 0.42. Hence C = 10.5 J/mole  K .
(c) We calculate © from the datum tha = 0.75 at T = 125 K .
The xcoordinate corresponding to 0.75 is 0.40. Hence
Now
So
Q.192. Demonstrate that molar heat capacity of a crystal at a temperature T << , where is the Debye temperature, is defined by Eq. (6.4f).
Ans. We use the formula (6.4d)
In the limit the third term in the bracket is exponentially small together wil derivatives.
Then we can drop the last term
Thus
Now from the table in the book
Thus
Note Call the 3^{rd} term in the bracket above  U_{3} Then
The maximum value of is a finite + v e quantity C_{0} for 0 ≤ x < Thus
we see that U_{3 }is exponentially small as
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 . This is also a test of the “ lowness ” of the temperature We see that
Thus T^{3} law is obeyed and T_{1}, T_{2 }can we regarded low .
Q.194. Calculate the mean zeropoint energy per one oscillator of a crystal in terms of the Debye theory if the Debye temperature of the crystal is equal to .
Ans. The total zero point energy of 1 mole of the solid is Dividing this by the number of modes 3N we get the average zero point energy per mode. It is
Q.195. Draw the vibration energy of a crystal as a function of frequency (neglecting the zeropoint vibrations). Consider two cases: T = /2 and T = /4, where is the Debye temperature.
Ans. In the Debye model
Then (Total no of modes is 3N)
Thus
wc get ignoring zero point energy
Thus for
For
Plotting 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
On substituting  330 K .
To get the corresponding value of the maximum momentum we must know the dispersion relation . For small we know where v is velocity o f sound in the crystal. For an order of magnitude estimate we continue to use this result for high Then we estimate v from the values of the modulus of elasticity and density
We write
Then
Hence
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 T_{max} is known.
Ans. (a) From the formula
the maximum value E_{max }of E is determined in terms of n by
or
(b) Mean K.E. < E >is
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
Q.199. Find the number of free electrons per one sodium atom at T = 0 if the Fermi level is equal to E_{F} = 3.07 eV and the density of sodium is 0.97 g/cm^{3}.
Ans. We calculate the concentration n of electron in the Na metal from
we get from
From this we get the number of electrons per one Na atom as
where p = density of Na, M = molar weight in gm of Na, N_{A} = 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 . This must equal Thus
We calculate E_{F }first For Cu
Then E_{F = }7.01 eV
and T = 3 2 5 x 10^{4} K
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