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 $\frac{x}{π} \times 10^{-17} eVs$. The value of x __________ is (use $h=4.14 \times 10^{-15} eVs$, $c=3 \times 108 ms^{-1}$).

Answer 1

The correct answer is 828.
In the ground state energy =−13.6eV
So energy
n2−13.6eV​=−13.6+12.75
n2−13.6eV​=−0.85
n=16​
n=4
Angular momentum =2πnh​=2π4h​=π2h​
Angular momentum =π2​×4.14×10−15
=π828×10−17​eVs