Question:
Electromagnetic radiation of wavelength 663 $\mathrm{nm}$ is just sufficient to ionise the atom of metal A. The ionization energy of metal $\mathrm{A}$ in $\mathrm{kJ} \mathrm{mol}^{-1}$ is__________ (Rounded-off to the nearest integer)
${\left[\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}, \mathrm{c}=3.00 \times 10^{8} \mathrm{~ms}^{-1}\right.}$
$\left.\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}\right]$
Solution:
$\mathrm{E}=\frac{\mathrm{hc}}{\lambda} \times \frac{\mathrm{N}_{\mathrm{A}}}{1000}$
$=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8} \times 6.02 \times 10^{23}}{663 \times 10^{-9} \times 1000}$
$=3 \times 6.02 \times 10 \mathrm{~kJ}$
$=180.6 \mathrm{~kJ}$