Question

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

Answer 1

$E=\frac{hc}{\lambda }=\frac{(6.626\times {10}^{-34}Js)(3.0\times {10}^{8}m{s}^{-1})}{150\times {10}^{-12}m}=1.3252\times {10}^{-15}J=13.252\times {10}^{-16}J$
Energy of the electron ejected (K.E)
=$\frac{1}{2}{m}_e{v}^2=\frac{1}{2}(9.10939\times {10}^{-31}kg)(1.5\times {10}^{7}m{s}^{-1}{)}^2=10.2480\times {10}^{-17}J=1.025\times {10}^{-16}J$
Hence, the energy with which the electron is bound to the nucleus can be obtained as:
= E – K.E
= 13.252 × ${10}^{–16}$ J – 1.025 × ${10}^{–16}$ J
= 12.227 × $10–16$ J
$\frac{12.227\times {10}^{-16}}{1.602\times {10}^{-19}}eV=7.6\times {10}^{3}eV$