Question

Electromagnetic radiation of wavelength $663nm$ is just sufficient to ionise the atom of metal $A$. The ionization energy of metal $A$ in $kJmo{l}^{-1}$ 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

Ionisation energy of an atom of metal
A = Quantum energy of radiation of wavelength $663nm$
$=\frac{6.63\times {10}^{-34}\times 3\times {10}^8\times 6.02\times {10}^{23}}{663\times {10}^{-9}\times 1000}J=3\times {10}^{-19}J$
Ionisation energy per mol
$=3\times {10}^{-19}\times {10}^{-3}\times 6.02\times {10}^{23}$
$=180.6kJmo{l}^{-1}=181$ $kJmo{l}^{-1}$