Question

If a photon of wavelength 100 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 nucleus.

• A. $-11.76\times {10}^{9}eV$
• B. $11.76\times {10}^{3}J$
• C. $11.76\times {10}^{3}eV$
• D. $-11.76\times {10}^{-3}J$

Answer 1

The correct option is C $11.76\times {10}^{3}eV$

Energy of incident photon
$E=\frac{hc}{\lambda }$
$=\frac{6.626\times {10}^{-34}Js\times 3.0\times {10}^{8}m{s}^{-1}}{100\times {10}^{-12}m}$
$=1.988\times {10}^{-15}J$
Energy of ejected electron
$=\frac{1}{2}m{v}^2$
$=\frac{1}{2}\times 9.11\times {10}^{-31}kg\times (1.5\times {10}^{7}m{s}^{-1}{)}^2$
= $1.025\times {10}^{-16}J$

Energy with which the electron was bound to the nucleus
$=19.88\times {10}^{-16}–1.025\times {10}^{-16}=18.855\times {10}^{-16}J$
Converting the value to eV
$=\frac{18.855\times {10}^{-16}}{1.602\times {10}^{-19}}=11.76\times {10}^{3}eV$