An electron in the ground state of a hydrogen atom absorbs 12.09 eV energy. The angular momentum of the electron increases by

The correct option is 2(h/2π), which corresponds to the second option:
Option 2: 2(h/2π)

Step-by-step Explanation:

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

2. Calculate the energy of the ground state (n = 1):
For the ground state, $n=1$:
$E_1=-\frac{13.6}{1^2}=-13.6\text{ eV}$

3. Find the energy of the excited state after absorption:
The electron absorbs 12.09 eV of energy. Therefore, the energy of the new state $E_n$ is:
$E_n=E_1+\Delta E$
$E_n=-13.6+12.09=-1.51\text{ eV}$

4. Identify the principal quantum number (n) of the excited state:
Using the energy formula for the new state:
$-1.51=-\frac{13.6}{n^2}$
$n^2=\frac{13.6}{1.51}\approx 9$
Taking the square root, we get:
$n=3$
So, the electron is excited from the ground state ($n_i=1$) to the third energy level ($n_f=3$).

5. Calculate the increase in angular momentum:
According to Bohr's second postulate, the angular momentum ($L$) of an electron in the $n$-th orbit is quantized and given by:
$L=\frac{nh}{2\pi }$
The change in angular momentum ($\Delta L$) when transitioning from $n_i=1$ to $n_f=3$ is:
$\Delta L=L_f-L_i$
$\Delta L=\frac{n_{f}h}{2\pi }-\frac{n_{i}h}{2\pi }$
$\Delta L=(n_f-n_i)\frac{h}{2\pi }$
Substituting the values $n_i=1$ and $n_f=3$:
$\Delta L=(3-1)\frac{h}{2\pi }=2\left(\frac{h}{2\pi }\right)$

Therefore, the angular momentum of the electron increases by 2(h/2π).