Five identifiable particles are distributed in three non-degenerate le

1.	Five identifiable particles are distributed in three non-degenerate levels with energies 0, E and 2E. The most probable distribution for total energy, 3E corresponds to what combination of options given below?1. N1 = 1, N2 = 2, N3 = 32. N1 = 3, N2 = 1, N3 = 13. N1 = 2, N2 = 3, N3 = 14. None
Answer» 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

Five identifiable particles are distributed in three non-degenerate levels with energies 0, E and 2E. The most probable distribution for total energy, 3E corresponds to what combination of options given below?1. N1 = 1, N2 = 2, N3 = 32. N1 = 3, N2 = 1, N3 = 13. N1 = 2, N2 = 3, N3 = 14. None

Answer» 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