Question

A light of energy $12.75 \,eV$ 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 $h=4.14 \times 10^{-15} eVs , c=3 \times 10^8 ms ^{-1}$ ).

Answer 1

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