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

Correct Answer: 828

\[ \text{Energy in ground state} = -13.6 \text{ eV} \] \[ - \frac{13.6}{n^2} = -13.6 + 12.75 \] \[ - \frac{13.6}{n^2} = -0.85 \] \[ n = \sqrt{16} \] \[ n = 4 \] \[ \text{Angular momentum} = \frac{nh}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi} \] \[ \text{Angular momentum} = \frac{2}{\pi} \times 4.14 \times 10^{-15} \] \[ = \frac{828 \times 10^{-17}}{\pi} \text{ eVs} \]