# [Statistical Physics] Probability of finding # photons in the mode

Flucky

## Homework Statement

A cavity contains black body radiation at temperature T = 500 K. Consider an optical mode in the cavity with frequency ω=2.5x10$^{13}$ Hz. Calculate

a) the probability of finding 0 photons in the mode
b) the probability of finding 1 photon in the mode
c) the mean number of photons in the mode.

## Homework Equations

Possibly <n> = $\frac{1}{exp(\frac{\hbar \omega}{k_{b} T})}$

## The Attempt at a Solution

Plugging the numbers into the equation above gives the answer to c (I think), which comes out to 1.855. However I thought that you could only have 0 or 1 photons in a given mode.

Not sure how to go about a) and b).

dauto
Photons are bosons. Bosons do not follow Pauli's exclusion principle. Any number (from zero to infinity) of photons may occupy any given mode.

Flucky
Ah ok, so the mean number of photons might still be ok.

Homework Helper
Just out of curiosity: how can you get <n> > 1 if all the factors in the exponential are > 0 ?

Flucky
You've just made me realize the equation should have a -1 on the bottom. Can't edit my original post for some reason.

Flucky
Ok think I've got the relevant equation for a) and b) now.

P(n) = $\frac{1 - exp(-\frac{ħw}{kT})}{exp(\frac{nħw}{kT})}$

where n is the number of photons in the mode.

Last edited:
Homework Helper
Gold Member
Ok think I've got the relevant equation for a) and b) now.

P(n) = $\frac{1 - exp(-\frac{ħw}{kT})}{exp(-\frac{nħw}{kT})}$

where n is the number of photons in the mode.

Note that in your expression P(n) → ∞ as n→∞. So, it can't be correct since a probability can't be greater than 1.

Maybe you need to switch the numerator and denominator.

Flucky
Ooh thanks for pointing that out, I accidentally put a minus in there.