### Atomic Physics Question 10

#### Question 10 - 01 February - Shift 1

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 $h=4.14 \times 10^{-15} eVs, c=3 \times 10^{8}$ $ms^{-1}$ ).

## Show Answer

#### Answer: (828)

#### Solution:

#### Formula: Angular momentum

In the ground state energy $=-13.6 eV$

So energy

$ \begin{aligned} & \frac{-13.6 eV}{n^{2}}=-13.6+12.75 \\ & \frac{-13.6 eV}{n^{2}}=-0.85 \\ & n=\sqrt{16} \\ & n=4 \end{aligned} $

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 $