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

$2(h/2\pi)$

The correct answer is Option (2) → $2(h/2\pi)$

Energy levels of hydrogen atom are given by:

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

Ground state (n = 1): $E_1 = -13.6\ \text{eV}$

After absorbing 12.09 eV, total energy = $E_1 + 12.09 = -13.6 + 12.09 = -1.51\ \text{eV}$

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

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

$n = 3$

Angular momentum $L = n \frac{h}{2\pi}$

Change in angular momentum = $(3 - 1)\frac{h}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi}$

∴ Increase in angular momentum = $\frac{h}{\pi}$