Statistical Physics NAT Level – 2 - Physics MCQ

The mean pressure of the perfect Fermi gas can be written as

The average energy of the system is

For an ideal fermion is a cubical box, the energy of single particle states are given by

If there is no interaction 
E_i = energy of i^th microstate of the system.



The correct answer is: 0.667

w = 10 kJ
q = –2 kJ
According to the first law of thermodynamics, ∆U = q + w
= –2 + 10 = 8 kJ

Systems following Fermi-Dirac statistics obey Pauli’s exclusion principal and each phase cell corresponding to one energy level which can be occupied by only 1 particle.
The correct answer is: 1

Since, the levels are non-degenerate there is only one state associated with each energy.
Let the number of particles is 3 energy state be N₁ , N₂ and N₃ respectively.
where N₁ + N₂ + N₃ = 6
As the particles are distinguishable, the number of microstates i.e. the number of ways of choosing N₁ , N₂ and N₃ particles from 6 particles is

W →  thermodynamic probability. It is maximum when N₁ !N₂ !N₃ ! is a minimum
where N₁ = N₂ = N₃ = 2
The corresponding total energy of distribution is

= 6E
The correct answer is: 6

For photons, B-E statistics is applicable
For photons,   

the density of states for photons between p and p + dp is

Since two independent direction of polarization

hv = dE
⇒ hdv = dE


Average number of photons 




The correct answer is: 3

The total number of cells upto energy level

Thus, the number of cells corresponding to energy interval E and E + dE is given by

The total number of cells between energy level 0 and E_F at T = 0, are equal to the total number of particles i.e.

Thus,  
E_fo is value of E_F at T = 0.
The average kinetic energy of the particle at T = 0 is given by

The correct answer is: 0.6

The total number of way W, of distributing 2 identical particles in the given system is

n_i = 2, g_i = 2

The correct answer is: 6

Probability of occupancy of a given state with energy E is given by the FermiDirac distribution formula which is


f(E) = 0.035
The correct answer is: 0.035

The Fermi energy of the electron gas is given by

with g_s = 2 

The correct answer is: 0.667

For B-E statististic, the particles are indistinguishable

So,  
The correct answer is: 1