The ionization enthalpy of hydrogen atom is $$1.312 \times 10^6\ J\ mol^{-1}$$. The energy required to excite the electron in the atom from $$n = 1$$ to $$n = 2$$ is

For a H-atom the energy of the electron in the $$n^{\text{th}}$$ orbit is given by Bohr’s expression
$$E_n = -\dfrac{R_H}{n^2}$$, where $$R_H$$ (in energy units per mole) is the Rydberg constant.

The given ionization enthalpy equals the energy needed to lift the electron from the ground state $$n = 1$$ to $$n = \infty$$, i.e.
$$\lvert E_1 \rvert = R_H = 1.312 \times 10^6\; \text{J mol}^{-1}$$.

Energy in the first excited state $$n = 2$$ is
$$E_2 = -\dfrac{R_H}{2^2} = -\dfrac{R_H}{4}$$.

The energy required to excite the electron from $$n = 1$$ to $$n = 2$$ is the difference
$$\Delta E = E_2 - E_1 = \left(-\dfrac{R_H}{4}\right) - \left(-R_H\right) = R_H\!\left(1 - \dfrac{1}{4}\right) = \dfrac{3}{4}R_H.$$

Substituting $$R_H = 1.312 \times 10^6\; \text{J mol}^{-1}$$:
$$\Delta E = \dfrac{3}{4} \times 1.312 \times 10^6 = 0.984 \times 10^6\; \text{J mol}^{-1} = 9.84 \times 10^5\; \text{J mol}^{-1}.$$

Hence, the required excitation energy is $$9.84 \times 10^5\ \text{J mol}^{-1}$$.

Option D which is: $$9.84 \times 10^5\ J\ mol^{-1}$$