Question:
The distribution function \( f(E) \) for a photon gas is given by:
The distribution function \( f(E) \) for a photon gas is given by:
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Photons obey Bose-Einstein distribution because they are bosons with zero chemical potential.
Updated On: Mar 26, 2025
- \( e^{-E / k_B T} \)
- \( \frac{E}{k_B T} e^{-E / k_B T} \)
- \( \frac{1}{e^{E / k_B T} - 1} \)
- \( \frac{1}{e^{-E / k_B T} + 1} \)
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The Correct Option is C
Solution and Explanation
Photon gas follows Bose-Einstein statistics. The occupation number distribution for bosons (such as photons) is given by:
\[ f(E) = \frac{1}{e^{E / k_B T} - 1} \]
\[ f(E) = \frac{1}{e^{E / k_B T} - 1} \]
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