A light of energy 12 . 75eV is incident on a hydrogen atom in its...

Let the electron jumped to ${n}^{th}$ excited state.

In the ground state, energy $E = - 13 . 6eV$

So, using relation $12 . 75 = 13 . 6\left[\dfrac{1}{{1}^{2}} - \dfrac{1}{{n}^{2}}\right]$

$⇒ 0 . 9375 = \left[1 - \dfrac{1}{{n}^{2}}\right] ⇒ {n}^{2} = 16$

$⇒ n = \sqrt{16} = 4$

Now, angular momentum, $L = \dfrac{nh}{2\pi} = \dfrac{4h}{2\pi} = \dfrac{2h}{\pi}$

$= \dfrac{2}{\pi} \times 4 . 14 \times {10}^{- 15}$

$= \dfrac{828 \times {10}^{- 17}}{\pi}eVs$

Hence, the value of $x = 828$.