# Photoelectic cross-section question

1. May 19, 2008

### jsc314159

1. The problem statement, all variables and given/known data
Show that if we consider photoemission from the 1s state of a charge Z atom,
$$\sigma \propto Z^5}}$$, in the limit $$p_fa_0/Z\hbar >> 1.$$

2. Relevant equations

$$\sigma = \frac{128a_0^3\pi e^2p_f^3}{3m\hbar^3\omegac[1+p_f^2a_0^2/\hbar^2]^4}}$$

3. The attempt at a solution
Actually I know how to do this. $$a_0 -> a_0/Z$$. My question is why is this so? What is the physical interpretation/what does it mean?

jsc

2. May 19, 2008

### jsc314159

I think the reason $$a_0 -> a_0/Z$$ is because the Coulomb potential in the Hamiltonian changes from $$q^2/r$$ to $$Zq^2/r$$. Is that right?

jsc