Problems and Solutions on Thermodynamics and Statistical Mechanics

Statisticd Physicice 181

Solution:
Fluctuations in the motion of free electrons in the conductor give rise
to fluctuation currents. If the current passing through the inductor is I(t),

then the average energy of the inductor is w = -I2, where iz is the mean-

square current. According to the principle of equipartition of energy, we

L-
2
1
have w = - kT. Hence
2

2021

Energy probability.

Find and make careful sketch of the probability density, p(E), for the

energy E of a single atom in a classical non-interacting monatomic gas in

thermal equilibrium.

(MITI

Solution:

When the number of gas atoms is very large, we can represent the states

of the system by a continuous distribution. When the system reaches ther-

mal equilibrium, the probability of an atom having energy E is proportional

to exp(-E/kT), where E = p2/2m, p being the momentum of the atom.

So the probability of an atom lying between p and p + dp is

Aexp(-p2/2mkT)d3p.

From

A 1 exp(-p2/2rnkT)d3p = 1 ,

A = (2~mkT)-~f~.

we obtain

Therefore,