Question

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?

Short answer

As a result, tungsten's photon energy is 72.474 keV.

Step-by-step solution

Determine the formulas:

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

$$\mathbf{\Delta }\mathbf{E}\mathbf{=}{\mathbf{E}}_{\mathbf{i}}\mathbf{-}{\mathbf{E}}_{\mathbf{f}}$$

Here,

$$\mathrm{\Delta }E=$$Change in energy

$$\begin{array}{c}{E}_i=\mathrm{I}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}\mathrm{E}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}\\ E_f=\mathrm{F}\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{l}\mathrm{E}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}\end{array}$$

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

$$E=\frac{Z^2}{n^2}{E}_o$$.

Calculate the photon energy for tungsten

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

Thus, $$n_i=\mathrm{\infty }$$ and $$n_f=1$$.

Also, the atomic number of tungsten is $$Z=73$$.

Thus, change in energy is calculated as

$$\mathrm{\Delta }E=\frac{Z^2}{n_i^2}{E}_o-\frac{Z^2}{n_{r^2}}{E}_o$$

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

Hence, the photon energy for tungsten is 72.474 keV.