1. Energy levels of hydrogen atom

   The energy of the electron in the $n^{\text{th}}$ orbit is $E_n = -\frac{13.6}{n^2}\ \text{eV}$

   Initially, the atom is in the ground state, so $E_1 = -13.6\ \text{eV}$

2. Energy absorbed

   The atom absorbs $12.09\ \text{eV}$, so the final energy becomes $E_f = -13.6 + 12.09 = -1.51\ \text{eV}$

3. Find the final orbit

   Using $E_n = -\frac{13.6}{n^2} = -1.51$

   $\frac{13.6}{n^2} = 1.51$

   $n^2 = \frac{13.6}{1.51} \approx 9$

   $n = 3$

   So the electron jumps from $n=1$ to $n=3$.

4. Orbital angular momentum in Bohr model

   In the $n^{\text{th}}$ orbit, $L_n = n\hbar$

   Initial angular momentum: $L_1 = 1\hbar$

   Final angular momentum: $L_3 = 3\hbar$

   Increase in angular momentum: $\Delta L = L_3 - L_1 = 3\hbar - \hbar = 2\hbar$

5. Numerical value

   $\hbar = 1.055 \times 10^{-34}\ \text{J s}$

   Therefore, $\Delta L = 2\hbar = 2(1.055 \times 10^{-34})$ $\Delta L = 2.11 \times 10^{-34}\ \text{J s}$

6. Option check

   • A: $1.05 \times 10^{-34}$ Js $= \hbar$ ❌
   • B: $2.11 \times 10^{-34}$ Js $= 2\hbar$ ✅
   • C: $3.16 \times 10^{-34}$ Js $= 3\hbar$ ❌
   • D: $4.22 \times 10^{-34}$ Js $= 4\hbar$ ❌

Hence, the correct answer is B.