Question

Electromagnetic radiation of wavelength 663 $nm$ is just sufficient to ionise the atom of metal A. The ionization energy of metal A in kJ $mo{l}^{-1}$ is _____.(Rounded-off to the nearest integer)
$[h=6.63\times 10^{-34}Js,c=3.00\times 10^{8}m{s}^{-1}$
$N_A=6.02\times 10^{23}mo{l}^{-1}]$

Answer 1

Answer: 180

$E=\frac{hc}{\lambda }\times \frac{N_A}{1000}$

$=\frac{6.63\times 10^{-34}\times 3\times 10^8\times 6.02\times 10^{23}}{663\times 10^{-9}\times 1000}$

$=3\times 6.02\times 10kJ$

$=180.6kJ$