If a photon of wavelength $150 \, pm$ strikes an atom and one of its inner bound electrons is ejected with a velocity of $1.5 \times 10^{7} \, ms^{-1}$, calculate the energy with which it is bound to the nucleus.

(N/A) The energy of the incident photon $(E)$ is given by $E = \frac{hc}{\lambda}$.
$E = \frac{(6.626 \times 10^{-34} \, Js)(3.0 \times 10^{8} \, ms^{-1})}{150 \times 10^{-12} \, m} = 1.3252 \times 10^{-15} \, J = 13.252 \times 10^{-16} \, J$.
The kinetic energy of the ejected electron $(K.E)$ is given by $K.E = \frac{1}{2} m_{e} v^{2}$.
$K.E = \frac{1}{2} (9.109 \times 10^{-31} \, kg)(1.5 \times 10^{7} \, ms^{-1})^{2} = 1.0248 \times 10^{-16} \, J$.
The binding energy $(B.E)$ is the difference between the incident photon energy and the kinetic energy of the ejected electron:
$B.E = E - K.E = 13.252 \times 10^{-16} \, J - 1.0248 \times 10^{-16} \, J = 12.2272 \times 10^{-16} \, J$.
Converting to $eV$:
$B.E = \frac{12.2272 \times 10^{-16} \, J}{1.602 \times 10^{-19} \, J/eV} \approx 7632 \, eV \approx 7.63 \times 10^{3} \, eV$.