Correct Answer – Option 2 : N1 = 3, N2 = 1, N₃ = 1

Explanation:

As the levels are non-degenerate, there is only one state for each energy.

Let the number of particles occupying the 3 energy states be N1, N2, and N3 respectively

Where 

N₁ + N₂ + N₃ = 5
 

The particles are identifiable, the number of ways of choosing the particles is
\(W = \frac{{5!}}{{{N_1}!\;{N_2}!{N_3}!}}\)

The energy of the system is

0 × N₁ + E × N₂ + 2 E × N₃ = 3E (given)

N2 + 2 N3 = 3  —–(1)
 

Now, the most probable distribution is the one in which W is a maximum, subject to constraint given by equation (1)

Thus if

N₂ = 1

N₃ = (3 -1) /2 = 1

and

N₁ = 5 – (N₁ + N₂) = 5 – 2 = 3

If N2 = 3, N3 = 0

and N1 = 5 – (3 + 0) = 2

No other distribution are possible

For

N₁ = 3, N₃ = 1 and N₃ = 1  

\(W = \frac{{5!}}{{3!}} = 20\)

So must probable distribution is 

N₁ = 3, N₂ = 1 and N₃ = 1

The correct answers are: 

N1 = 3, N2 = 1 and N3 = 1