The maximum characteristic x-ray photon energy comes from the capture of a free electron into a K shell vacancy. What is this photon energy in keV for tungsten, assuming the free electron has no initial kinetic energy?

Consider the formula for the change in energy from one level to other as follows:

\[{\bf{\Delta E = }}{{\bf{E}}_{\bf{i}}}{\bf{ - }}{{\bf{E}}_{\bf{f}}}\]

Here,

\[\Delta E = \]Change in energy

\[\begin{array}{c}{E_i} = {\rm{Initial Energy }}\\{E_f} = {\rm{Final Energy}}\end{array}\]

In addition, the equation for determining the energy of an electron in a hydrogen-like atom is as follows:

\[E = \frac{{{Z^2}}}{{{n^2}}}{E_o}\].

Consider the electron is transitioned from infinity to the ground state.

Thus, \[{n_i} = \infty \] and \[{n_f} = 1\].

Also, the atomic number of tungsten is \[Z = 73\].

Thus, change in energy is calculated as

\[\Delta E = \frac{{{Z^2}}}{{n_i^2}}{E_o} - \frac{{{Z^2}}}{{{n_{{r^2}}}}}{E_o}\]

Since the free electrons has no initial kinetic energy, we get

Hence, the photon energy for tungsten is 72.474 keV.