X Rays and the Ordering of the Elements — Question 5

Question 5

The K absorption edge lies above the K-alpha emission line

An atom has K and L binding energies $B_K=10.0\,\mathrm{keV}$ and $B_L=1.5\,\mathrm{keV}$. Ignore fine structure, recoil, and rearrangement. An incident photon of energy $12.0\,\mathrm{keV}$ ejects a K electron.

1. Find the K-edge threshold and outgoing photoelectron kinetic energy.

2. Find the emitted $K_\alpha$ energy when an L electron fills the hole, and account for the remaining energy stored in the final ion.

Question 5 - Solution

The K absorption edge lies above the K-alpha emission line

(a) Method and calculation

$$\boxed{E_{\mathrm{edge}}=B_K=10.0\,\mathrm{keV}}.$$$$\boxed{K_e=E_{\mathrm{in}}-B_K=2.0\,\mathrm{keV}}.$$

(b) Interpretation and further calculation

$$\boxed{E_{K\alpha}=B_K-B_L=8.5\,\mathrm{keV}}.$$$$E_{\mathrm{in}}=K_e+E_{K\alpha}+B_L=2.0+8.5+1.5=\boxed{12.0\,\mathrm{keV}}.$$

Physical check

The final ion has an L vacancy and retains ionization energy. Confusing the edge with the emission line loses that residual energy in the accounting.