[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.

How do I go about answering a and b?

 

[BvU]

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.

For any future readers:

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

where n is the number of photons in the mode.

 

[TSny]

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

For any future readers:

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.