1. Energy levels of hydrogen atom

For a hydrogen atom, the energy of the electron in the $n^{\text{th}}$ orbit is

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

Here, $13.6\,\text{eV}$ is the ionization energy from the ground state ($n=1$).

2. Energy of electron at $n=2$

Substitute $n=2$:

$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4\,\text{eV}$

3. Energy required to remove electron from $n=2$

To ionize the atom from $n=2$, the electron must be taken from $E_2 = -3.4\,\text{eV}$ to $0\,\text{eV}$.

So required energy is

$\Delta E = 0 - (-3.4) = 3.4\,\text{eV}$

4. Checking options

• A: $10.2\,\text{eV}$ — incorrect
• B: $0\,\text{eV}$ — incorrect
• C: $3.4\,\text{eV}$ — correct
• D: $6.8\,\text{eV}$ — incorrect

Therefore, the correct answer is Option C.