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 $1.5\times {10}^{7}m{s}^{-1}$, calculate the energy with which it is bound to the nucleus.

Answer 1

The energy of the photon,

$E=\frac{hc}{\lambda }$

$=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{150\times {10}^{-12}}$

$=1.325\times {10}^{-15}J$.

The energy of the ejected electron is

$\frac{1}{2}m{v}^2=\frac{1}{2}\times 9.11\times {10}^{-31}\times (1.5\times {10}^7{)}^2=1.025\times {10}^{-16}$.

The energy with which the electron was bound to the nucleus is:

Energy of photon $-$ Energy ejected by electron $=13.25\times {10}^{-16}-1.025\times {10}^{-16}=12.225\times {10}^{-16}J$

$=7.63\times {10}^{3}eV$.