1. Identify the transition

For hydrogen atom, the energy of the $n$th orbit is

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

Ground state means $n=1$, so

$E_1 = -13.6\,\text{eV}$

If the atom absorbs $10.2\,\text{eV}$, then the final energy becomes

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

Now,

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

$\frac{13.6}{n^2} = 3.4$

$n^2 = 4 \implies n=2$

So the electron goes from $n=1$ to $n=2$.

2. Use Bohr's angular momentum formula

According to Bohr's quantization,

$L_n = n\frac{h}{2\pi} = n\hbar$

Thus,

• Initial angular momentum: $L_1 = \frac{h}{2\pi}$

• Final angular momentum: $L_2 = 2\frac{h}{2\pi}$

Hence increase in angular momentum is

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

3. Substitute given value of $h$

Given,

$h = 6.6 \times 10^{-34}\,\text{Js}$

Therefore,

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

Using $\pi \approx 3.14$,

$\Delta L \approx \frac{6.6 \times 10^{-34}}{6.28}$

$\Delta L \approx 1.05 \times 10^{-34}\,\text{Js}$

4. Match with options

$\boxed{1.05 \times 10^{-34}\,\text{Js}}$

So the correct option is B.