Answer to Question #169528 in Molecular Physics | Thermodynamics for Khushi

Question

For a Bose-Einstein system, the expression for the thermodynamic probability is

W=(3.14)((g_i+N_i-1)!/(N_i!(g_i-1)!))

Derive an expression for the Bose-Einstein distribution function.

Answer 1

$$f=(e^{\frac {E-\mu _\epsilon}{k \tau}}-1)^{-1}$$

$$\mu_\epsilon=\frac {dU}{dn}=0$$

$$f=(e^{\frac {h\nu}{kT}} -1)^{-1}$$