# probability of emission vs absorption in thermal radiation

I am trying to make sense of the thermal radiation emitted by a gas.

The radiative transfer equation is $$\frac{dI}{dx} = \epsilon - \kappa I,$$ where I is the intensity, $$\epsilon$$ is the emissivity coefficient and $$\kappa$$ is the absorption coefficient. If there is no incident radiation on the gas, the solution is $$I = \frac{\epsilon}{\kappa}(1 - e^{-\kappa x}).$$

For the intensity to agree with Planck's law, we must have $$\frac{\epsilon}{\kappa} = \frac{2hf^3}{c^2}\frac{1}{e^{hf/kT} -1}.$$

Now let's imagine that the gas has only two energy levels. We could rewrite the last equation like this:

$$\frac{\epsilon}{\kappa} = \frac{2hf^3}{c^2}\frac{Ne^{-hf/kT}}{N(1-e^{-hf/kT})} = \frac{2hf^3}{c^2}\frac{N_{2}}{N_{1}},$$ where $$N_{1}$$ and $$N_{2}$$ are the number of atoms in energy levels 1 and 2 (Boltzmann).

Therefore, the ratio of emissivity and absorption coefficients is proportional to the ratio of number of atoms in levels 2 and 1, which makes sense since an atom has to be in level 2 to emit and level 1 to absorb.

But what about the fraction $$\frac{2hf^3}{c^2}$$? Why would emission be more probable (with respect to absorption) for higher values of $$f$$, i.e. for energy levels that are farther apart (assuming they are equally populated)?

• – eranreches Jun 5 at 5:02
• Your solution has zero intensity at $x=0$ and finite intensity in infinite space $x>0$. This function describes quite an unphysical situation. What is the original problem? – Ján Lalinský Jun 5 at 16:50