Hello!(adsbygoogle = window.adsbygoogle || []).push({});

Let's consider again a system of atoms with only two permitted energy levels [itex]E_1[/itex] and [itex]E_2 > E_1[/itex]. When electrons decay from [itex]E_2[/itex] level to [itex]E_1[/itex], they generate a photon of energy [itex]E_{21} = E_2 - E_1 = h \nu[/itex]. The number of photons (per unit frequency, per unit volume) emitted by such a system in thermal equilibrium at a temperature [itex]T[/itex] can be determined dividing its black body radiation [itex]\rho (\nu)[/itex] by [itex]h \nu[/itex]:

[itex]

n_{ph} (E_{21}) = \rho (\nu) \displaystyle \frac{1}{h \nu} = \displaystyle \frac{8 \pi h \nu^3}{c^3 \left( e^{h \nu / (k_B T)} - 1 \right)} \frac{1}{h \nu} = \frac{8 \pi \nu^2}{c^3 \left( e^{h \nu / (k_B T)} - 1 \right)} = \frac{8 \pi E_{21}^2}{h^2 c^3 \left( e^{h \nu / (k_B T)} - 1 \right)}

[/itex]

where [itex]h[/itex] is the Planck constant, [itex]k_B[/itex] is the Boltzmann constant, [itex]c[/itex] is the speed of light. This computation is about spontaneous and stimulated emission in a LASER system.

In this document (page 10, formula (2.13)), [itex]n_{ph} (E_{21})[/itex] is slightly different. It is

[itex]

n_{ph} (E_{21}) = \displaystyle \frac{8 \pi n_r^3 E_{21}^2}{h^3 c^3 \left( e^{h \nu / (k_B T)} - 1 \right)}

[/itex]

So:

1) If the above computation is correct, why in the book the denominator contains [itex]h^3[/itex] instead of [itex]h^2[/itex]?

2) What can the refractive index be related to? Maybe is it due to the fact that the material is a semiconductor and not the vacuum?

Thank you anyway!

Emily

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# I Planck formula and density of photons

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**