Question

A light of energy $12.75eV$ is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is $\frac{x}{\pi }\times 10^{-17}eVs$. The value of $x$ is ___________(use $h=4.14\times 10^{-15}eVs,c=3\times 10^{8}m{s}^{-1}$ ).

Answer 1

Answer: 828

In the ground state energy $=-13.6eV$
So energy
$\frac{-13.6eV}{n^2}=-13.6+12.75$
$\frac{-13.6eV}{n^2}=-0.85$
$n=\sqrt{16}$
$n=4$
Angular momentum $=\frac{nh}{2\pi }=\frac{4h}{2\pi }=\frac{2h}{\pi }$
Angular momentum $=\frac{2}{\pi }\times 4.14\times 10^{-15}$
$=\frac{828\times 10^{-17}}{\pi }eVs$