he ionization enthalpy of hydrogen atom is 1.312 × 10^6 J mol^(-1). The energy required to excite the electron in the atom from n = 1 to n = 2 is

The correct option is 9.84 × 10⁵ J mol⁻¹.

To understand why this is the correct answer, let us break down the physical concepts and the mathematical calculations step-by-step.
The ionization enthalpy of a hydrogen atom is the energy required to remove an electron completely from its ground state (where the principal quantum number $n=1$) to an infinite distance from the nucleus (where $n=\infty$).
Thus, the ionization energy ($E_{\text{ionization}}$) is the difference in energy between these two states:

$E_{\text{ionization}}=E_{\infty }-E_1$

Since the energy of an electron at an infinite distance ($E_{\infty }$) is taken as zero, we have:

$E_{\text{ionization}}=0-E_1=-E_1$

Given that the ionization enthalpy of the hydrogen atom is $1.312\times {10}^6{\text{J mol}}^{-1}$, the energy of the electron in the ground state ($n=1$) is:

$E_1=-1.312\times {10}^6{\text{J mol}}^{-1}$

According to Bohr's model of the hydrogen atom, the energy of an orbit with principal quantum number $n$ is inversely proportional to $n^2$:

$E_n=\frac{E_1}{n^2}$

Using this relationship, we can calculate the energy of the first excited state ($n=2$):

$E_2=\frac{E_1}{2^2}=\frac{E_1}{4}$

Substitute the value of $E_1$ into the equation:

$E_2=\frac{-1.312\times {10}^6{\text{J mol}}^{-1}}{4}=-0.328\times {10}^6{\text{J mol}}^{-1}$

The energy required ($\Delta E$) to excite the electron from the ground state ($n=1$) to the first excited state ($n=2$) is the difference between their respective energy levels:

$\Delta E=E_2-E_1$

Substitute the values of $E_1$ and $E_2$ into the equation:

$\Delta E=\left(-0.328\times {10}^6\right)-\left(-1.312\times {10}^6\right){\text{J mol}}^{-1}$

Simplify the expression:

$\Delta E=\left(-0.328+1.312\right)\times {10}^6{\text{J mol}}^{-1}$

$\Delta E=0.984\times {10}^6{\text{J mol}}^{-1}$

Expressing this in scientific notation:

$\Delta E=9.84\times {10}^5{\text{J mol}}^{-1}$

This calculation matches the correct option exactly.