Photoionization by black-body radiation I am trying to solve the question:

"Show that if an atom is photoionized by black-body radiation with the
temperature $T^*$ such that $kT^*\ll I_H,$ the average kinetic energy acquired by an ejected electron is about $kT^*$."

I understand that the energy of ionizing photon should be equal to the energy differences of the levels in the atom+ kinetic energy of the electron, BUT, I don't understand how to refer to the energy of photons, should it be for each mode or total energy of the spectrum, I'm confused.
 A: The only photons absorbed are those with energies higher than the ionization energy, that is much higher than $k_BT^*$:
$$h\nu = I_H + K > I_H\gg kT^*.$$
In this limit, $h\nu\gg kT^*$ the Planck's law becomes Wien's law:
$$
I(\nu, T)=\frac{2h\nu^3}{c^2}e^{-\frac{h\nu}{kT^*}} = 
\frac{2}{h^2c^2}(I_H + K)^3e^{-\frac{I_H+K}{kT^*}}
$$
Since the exponent drops quickly, we could omit $K$ in the prefactor, giving us essentially an exponential distribution of kinetic energies with mean $kT^*$. (Instead of omitting $K$ in the prefactor, one could try to do some more precise math, but it would produce pretty much the same result. Don't forget to normalize the distribution though.)
A: Energy Balance: KE of electron = Energy of photon - I.
For all photons with Energy E>I.
I calculated the Energy of the photons above I dividend by the number of photons above I.
Energy of photons above I: Integral over x^3 exp(-x) dx
Number of photons above I: Integral over x^2 exp(-x) dx
Both integrale from I/kT to Infinity. x is the photon energy E/ kT. I skipped the -1 in the denominator of the distribution function because I>>kT.
The normalisation constants area, c, h, pi ... cancel. The result is in units of kT.
I solved it with Wolframalpha.
Result: I (in kT) + 1 + something very small. So the average energy above I is kT. Something small is in the order of 1/I.
Happy Newtonmas.
