When a hydrogen atom in the ground state absorbs 10.2 eV of energy,...

The energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by $E_n = -\dfrac{13.6}{n^2} \text{ eV}$.

For the ground state ($n=1$), $E_1 = -13.6 \text{ eV}$.

When the electron absorbs $10.2 \text{ eV}$ of energy, its new energy becomes:

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

Equating this to the energy expression for the $n^{\text{th}}$ orbit:

$-\dfrac{13.6}{n^2} = -3.4$

$n^2 = \dfrac{13.6}{3.4} = 4 \Rightarrow n = 2$

According to Bohr's model, the orbital angular momentum of an electron in the $n^{\text{th}}$ orbit is $L = \dfrac{nh}{2\pi}$.

The increase in orbital angular momentum is:

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

Substituting the values of $h = 6.63 \times 10^{-34} \text{ J s}$ and $\pi = 3.14$:

$\Delta L = \dfrac{6.63 \times 10^{-34}}{2 \times 3.14} \approx 1.05 \times 10^{-34} \text{ J s}$