The relativistic quantum electrodynamics is rigorously applied to calculate the cross section for photoelectric absorption of a Dirac electron in the field of a nucleus (of charge Ze). The case of quantum energies $h\nu$ greater than the proper energy $m{c}^2$ of the electron is considered. The total photo-absorption from the $K$-shell is found in this way to be approximately ${\tau }_K=1.16\mathrm{}\times \frac{{10}^{-23\mathrm{}}{Z}^5\lambda ·R}{{\alpha }^{2{\alpha }^2}{e}^{\alpha (\pi -2\mathrm{}\alpha )}},$ where $R\sim 1+\frac{7m{c}^2}{3h\nu }\to 1$, as $\nu \to \infty$, and $\alpha =\frac{2\pi e^{2}Z}{\mathrm{hc}}$. The factor depending on $\alpha$ is seen to modify the result appreciably for heavy elements, so that for lead, and the gamma-rays from Th C", ($Z=82\mathrm{}$, $\lambda =4.7\mathrm{}$ X.U.), ${\tau }_K=1.1\mathrm{}\times {10}^{-24\mathrm{}}$ ${\mathrm{cm}}^2$. The agreement with experiment is satisfactory.

Close

PrevNext

View figure in article