Question #169528

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

W=(3.14)((gi+Ni-1)!/(Ni!(gi-1)!))

Derive an expression for the Bose-Einstein distribution function.


Expert's answer

f=(eEμϵkτ1)1f=(e^{\frac {E-\mu _\epsilon}{k \tau}}-1)^{-1}

μϵ=dUdn=0\mu_\epsilon=\frac {dU}{dn}=0

f=(ehνkT1)1f=(e^{\frac {h\nu}{kT}} -1)^{-1}


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