Examveda
Examveda

A photon gas is at thermal equilibrium at temperature T. The mean number of photons in an energy state $$\varepsilon = h\omega $$  is

A. $$\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) + 1$$

B. $$\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) - 1$$

C. $${\left[ {\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) + 1} \right]^{ - 1}}$$

D. $${\left[ {\exp \left( {\frac{{\hbar \omega }}{{{k_B}T}}} \right) - 1} \right]^{ - 1}}$$

Answer: Option D


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Comments ( 1 )

  1. Sr Yadav
    Sr Yadav :
    9 months ago

    Answer please

Related Questions on Thermodynamics and Statistical Physics

A system has two energy levels with energies E and 2E. The lower level is four-fold degenerate while the upper level is doubly degenerate. If there are N non-interacting classical particles in the system, which is in thermodynamic equilibrium at temperature T, the fraction of particles in the upper level is

A. $$\frac{1}{{1 + {e^{ - \varepsilon /{k_B}T}}}}$$

B. $$\frac{1}{{1 + 2{e^{\varepsilon /{k_B}T}}}}$$

C. $$\frac{1}{{2{e^{\varepsilon /{k_B}T}} + 4{e^{2\varepsilon /{k_B}T}}}}$$

D. $$\frac{1}{{2{e^{\varepsilon /{k_B}T}} - 4{e^{2\varepsilon /{k_B}T}}}}$$