Statistical Physics NAT Level - 1 - IIT JAM MCQ

We know that E = U/N
and 
and 


The correct answer is: 0.6

We know that 
and 

v_mp = 0.8165 v_rms
The correct answer is: 0.8165

Since the particles which exhibit odd integral spins are fermions and hence they obey Pauli’s exclusion principles. According to Pauli’s exclusion principle, not more than one particle can exist in the one quantum state. Hence, the number of assessible states N = 20.
The correct answer is: 20

For a single particle, these are 3 degrees of position coordinates x, y, z and 3 degrees of momentum coordinates p_x, p_y, p_z, hence total coordinates = 6.
The correct answer is: 6

= 5eV
The correct answer is: 5

The probability of finding the system in the energy state with energy ε may be defined

But  hence probability of finding the sytem with ε = 0 is

In question, βε₀ = 2 so

P(0) = (1 + 2cosh 2)^-1
⇒ α = 0.
The correct answer is: -1

For a rigid diatomic molecule, there is no rotation possible along the inter-nuclear axis. That remove one angle from 3 Euler angles needed to describe the orientation of the molecules, and hence also removes an angular momentum component. That leaves 4 angular degrees of freedom plus 3 coordinate and 3 momenta for the center of mass motions.
∴ One diatomic molecule has phase space of dimension 10, so the number ofphase space for10 diatomic molecule = 10 ×10 = 100.
The correct answer is: 100

For a 2D moving in a plane as in case of a circular motion, we have two degrees of freedom for the q i.e. for x – y plane, q₁ = x and q₂ = y, and two degrees of freedom for the velocity given by u_x and v_y and thus p_x and p_y.
⇒ dimension of phase space = 2 + 2 = 4.
The correct answer is: 4

Since the particle is moving on two dimensional surface, its degree of freedom = 2.
Hence, for N non-interacting particles total degree of freedom = 2N
The energy per degree of freedom = 1/2 k_BT
Hence, total energy 
U = Nk_BT
The correct answer is: 1