91Ó°ÊÓ

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.