Question

Electromagnetic radiation of wavelength $663nm$ is just sufficient to ionize the atom of metal $A$ . The ionization energy of metal $A$ in $kJmol^{- 1}$ is ______.
(Rounded off to the nearest integer)
Use:
$h=6.63\times 10^{- 34}Jsc=3.00\times 10^{8}ms^{- 1}N_{A}=6.02\times 10^{23}mol^{- 1}$

Answer 1

Energy req. to ionize an atom of metal $A=\frac{hc}{\lambda }=\frac{hc}{663 nm}$
for $1mole$ atoms of $A$ Total energy required $=N_{A}\times \frac{hc}{\lambda }$
$=\frac{6 . 023 \times 10^{23} \times 6 . 63 \times 10^{- 34} \times 3 \times 10^{8}}{663 \times 10^{- 9}}$
$=6.023\times 3\times 10^{23 - 34 + 8 + 7}$
$=18.04\times 10^{4}J/mol$
$=180.4KJ/mol$
Nearest Integer $=180KJ/Mol$ .