For distinguishable molecules are distributed in energy level E1 and E...
Solution:
The number of ways to distribute n distinguishable molecules among g distinguishable energy levels with energy E1, E2, …, Eg is given by the formula:
W = (n!)/(n1!n2!...ng!)
where n1, n2, …, ng are the number of molecules in each energy level.
In this case, we have:
n = 5 molecules
g = 2 energy levels (E1 and E2)
E1 has degeneracy 4
E2 has degeneracy 5
n1 = 3 molecules in E1
n2 = 2 molecules in E2
Calculating the number of microstates:
Number of microstates = W1 * W2
where W1 is the number of ways to distribute 3 molecules among 4 degenerate energy levels, and W2 is the number of ways to distribute 2 molecules among 5 degenerate energy levels.
W1 = (3+4-1)!/(3!(4-1)!) = 20
W2 = (2+5-1)!/(2!(5-1)!) = 15
Number of microstates = W1 * W2 = 20 * 15 = 300
However, we need to consider the arrangement of the 5 molecules among themselves. This can be done in 5! ways. Therefore, the total number of microstates is:
Total number of microstates = 5! * 300 = 36,000
But, we have to divide by the number of ways to arrange the 3 molecules in E1 and the 2 molecules in E2. This can be done in 3! and 2! ways, respectively. Therefore, the final answer is:
Final answer = 36,000 / (3! * 2!) = 16,000
Therefore, the number of microstates with three molecules in energy level E1 and two in energy levels E2 is 16,000.