A hydrogen atom in its ground state absorbs $$10.2$$ eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of (Given, Planck's constant $$= 6.6 \times 10^{-34}$$ Js).

A hydrogen atom in the ground state (n = 1) absorbs 10.2 eV of energy. We need to find the increase in angular momentum.

The energy levels of hydrogen are given by:

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

Ground state energy: $$E_1 = -13.6$$ eV

After absorbing 10.2 eV:

$$E_f = -13.6 + 10.2 = -3.4 \text{ eV}$$

Since $$E_2 = -\frac{13.6}{4} = -3.4$$ eV, the electron transitions to n = 2.

The angular momentum in the nth orbit is:

$$L_n = \frac{nh}{2\pi}$$

Change in angular momentum:

$$\Delta L = L_2 - L_1 = \frac{2h}{2\pi} - \frac{h}{2\pi} = \frac{h}{2\pi}$$

Substituting $$h = 6.6 \times 10^{-34}$$ Js:

$$\Delta L = \frac{6.6 \times 10^{-34}}{2\pi} = \frac{6.6 \times 10^{-34}}{6.28} = 1.05 \times 10^{-34} \text{ Js}$$

The correct answer is Option B.