The excitation energy of a hydrogen atom from its ground state to its third excited state is

The correct option is 12.75 eV.

To find the excitation energy of a hydrogen atom from its ground state to its third excited state, we use the energy formula for a hydrogen-like atom.

The energy of an electron in the n^th energy level of a hydrogen atom is given by:

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

Here, n represents the principal quantum number.

1. Ground State (n = 1):
The energy of the ground state is:

$E_1=-\frac{13.6}{1^2}=-13.6\text{ eV}$

2. Third Excited State (n = 4):
The excited states of a hydrogen atom are indexed as follows:
- First excited state: n = 2
- Second excited state: n = 3
- Third excited state: n = 4
The energy of the third excited state is:

$E_4=-\frac{13.6}{4^2}=-\frac{13.6}{16}=-0.85\text{ eV}$

3. Excitation Energy:
The excitation energy is the energy required to transition the electron from the lower energy level (ground state) to the higher energy level (third excited state):

$\Delta E=E_4-E_1$

$\Delta E=-0.85\text{ eV}-(-13.6\text{ eV})$

$\Delta E=13.6\text{ eV}-0.85\text{ eV}=12.75\text{ eV}$

Therefore, the excitation energy is 12.75 eV.