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

Energy levels of hydrogen atom are given by the formula: $$E_n = -\frac{13.6\text{ eV}}{n^2}$$.

The ground state energy is $$E_1 = -13.6\text{ eV}$$, and the photon energy absorbed is $$E_{\text{photon}} = 12.75\text{ eV}$$. The atom is excited from $$n=1$$ to some level $$n$$, so we write the energy difference equation:

$$E_n - E_1 = E_{\text{photon}}\quad\Longrightarrow\quad \left(-\frac{13.6}{n^2}\right) - (-13.6) = 12.75\quad\text{$$-(1)$$}$$

From equation $$(1)$$ we have:
$$-13.6\Bigl(\frac{1}{n^2}-1\Bigr) = 12.75\quad\Longrightarrow\quad 13.6\Bigl(1 - \frac{1}{n^2}\Bigr) = 12.75.$$

Dividing both sides by $$13.6$$ gives:
$$1 - \frac{1}{n^2} = \frac{12.75}{13.6} = 0.9375.$$

Hence, $$\frac{1}{n^2} = 1 - 0.9375 = 0.0625 \quad\Longrightarrow\quad n^2 = 16 \quad\Longrightarrow\quad n = 4.$$

In Bohr’s model, the electron’s angular momentum is quantized as: $$L = n\hbar = n\frac{h}{2\pi}\quad\text{$$-(2)$$}$$

Substitute $$n = 4$$ and $$h = 4.14\times 10^{-15}\text{ eVs}$$ into $$(2)$$:
$$L = 4\times\frac{4.14\times10^{-15}}{2\pi}\text{ eVs} = \frac{16.56\times10^{-15}}{2\pi}\text{ eVs} = \frac{8.28\times10^{-15}}{\pi}\text{ eVs}.$$

Write $$8.28\times10^{-15}\text{ eVs}$$ as $$828\times10^{-17}\text{ eVs}$$, so
$$L = \frac{828\times10^{-17}}{\pi}\text{ eVs} = \frac{x}{\pi}\times10^{-17}\text{ eVs}.$$

By comparison, the value of $$x$$ is 828.