Question

four distinguishable molecules are distributed in energy levels E1 and E2 with degeneracy 2 and 3,respectively. Calculate number of microstates with 3 molecules in energy level E1 and 1 in E2.
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Answer 1

Explanation:

96 is right option by using that formula when degeneracy is given

Answer 2

Answer:

The number of microstates with 3 molecules in energy level E₁ and 1 molecule in E₂ is equal to 96.

Explanation:

Given, the number of distinguishable molecules, $N=4$

The degeneracy of E₁ energy level, $g_{1}=2$

The degeneracy of E₂ energy level, $g_{2}=3$

Now, given the number of molecules in E₁ energy level $n_{1}=3$

The number of molecules in E₂ energy level, $n_{2}=1$

The general formula for the total number of microstates:

$W=N! \prod \frac{(g_{i})^{n_{i}}}{n_{i}!}$

We can write the modified formula for given data as:

$W=N\frac{(g_{1})^{n_{1}}}{n_{1}!}*\frac{(g_{2})^{n_{2}}}{n_{2}!}$                                              .......................(1)

Now put the values in equation (1);

$W=4!*(\frac{(2)^{3}}{3!}*\frac{(3)^{1}}{1!})$

$W=4*3*2*1(\frac{8}{3*2*1} *\frac{3}{1} )$

$W=96$

Therefore, the number of microstates for four distinguishable molecules in E₁ energy level and E₂ energy level is equal to 96.