Question

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

Answer 1

Energy required to ionize an atom of metal ${}^{\prime }{A}^{\prime }=\frac{hc}{\lambda }=\frac{hc}{663\text{nm}}$
For $1$ mole atoms of 'A'
Total energy required $=N_A\times \frac{hc}{\lambda }$
$=\frac{6.02\times {10}^{23}\times 6.63\times {10}^{-34}\times 3\times {10}^8}{663\times {10}^{-9}}$
$=6.02\times 3\times {10}^{23-34+8+7}$
$=18.06\times {10}^4\text{J/mol}$
$=180.4\text{kJ/mol}$
Nearest Integer value$=180\text{kJ/mol.}$