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, $$c = 3 \times 10^8$$ m s$$^{-1}$$)

We have a photon of energy 12.75 eV incident on a hydrogen atom in the ground state. The energy levels of hydrogen are given by

$$E_n = -\frac{13.6}{n^2} \text{ eV}$$

The ground state energy is $$E_1 = -13.6$$ eV. After absorbing 12.75 eV, the energy becomes

$$E = -13.6 + 12.75 = -0.85 \text{ eV}$$

Now finding the excited state,

$$-\frac{13.6}{n^2} = -0.85 \implies n^2 = \frac{13.6}{0.85} = 16 \implies n = 4$$

The angular momentum in the $$n$$-th orbit by Bohr's postulate is

$$L = \frac{nh}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi}$$

Substituting $$h = 4.14 \times 10^{-15}$$ eVs,

$$L = \frac{2 \times 4.14 \times 10^{-15}}{\pi} = \frac{8.28 \times 10^{-15}}{\pi} = \frac{828 \times 10^{-17}}{\pi} \text{ eVs}$$

Comparing with $$\frac{x}{\pi} \times 10^{-17}$$ eVs, we get $$x = 828$$. So, the answer is $$828$$.