1. Use Bohr energy levels for hydrogen

For hydrogen atom, $E_n = -\frac{13.6}{n^2}\text{ eV per atom}$

So the ionization energy from ground state $n=1$ corresponds to the magnitude of $E_1$.

Given in the question: $\text{Ionization enthalpy} = 1.312 \times 10^6\ \text{J mol}^{-1}$

This is the energy needed to take the electron from $n=1$ to $n=\infty$.

2. Find energy of levels in terms of ionization enthalpy

Since energy varies as $1/n^2$, $E_1 = -1.312 \times 10^6\ \text{J mol}^{-1}$ $E_2 = -\frac{1.312 \times 10^6}{4}\ \text{J mol}^{-1}$

Thus, $E_2 = -0.328 \times 10^6 = -3.28 \times 10^5\ \text{J mol}^{-1}$

3. Energy required for excitation from $n=1$ to $n=2$

Required energy, $\Delta E = E_2 - E_1$ $= \left(-\frac{1.312 \times 10^6}{4}\right) - \left(-1.312 \times 10^6\right)$ $= 1.312 \times 10^6\left(1 - \frac14\right)$ $= 1.312 \times 10^6 \times \frac34$ $= 0.984 \times 10^6$ $= 9.84 \times 10^5\ \text{J mol}^{-1}$

4. Match with options

$9.84 \times 10^5\ \text{J mol}^{-1}$ corresponds to Option D.

5. Comparison with stored answer

Stored correct answer: D

Derived answer: D

They match.