Question

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

Answer 1

Answer: 180

Energy req. to ionize an atom of metal $A=\frac{hc}{\lambda }=\frac{hc}{663nm}$
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 $=180KJ/Mol$ .