The ionization enthalpy of hydrogen atom is 1.312 × 106 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.

Step-by-Step Explanation:

1. Understand Ionization Enthalpy:
Ionization enthalpy ($E_{\infty }$) is the energy required to remove an electron completely from the ground state ($n=1$) to infinity ($n=\infty$).
The energy of the electron in the ground state ($E_1$) is the negative of the ionization energy:

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

2. Determine the Energy of the Second Orbit ($n=2$):
For a hydrogen atom, the energy of an electron in any orbit $n$ is given by:

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

Therefore, for the first excited state ($n=2$), the energy ($E_2$) is:

$E_2=\frac{-1.312\times {10}^6}{2^2}=\frac{-1.312\times {10}^6}{4}=-0.328\times {10}^6\text{ J mol}{-1}^{}$

3. Calculate the Energy Required for Excitation ($\Delta E$):
The energy required to excite the electron from $n=1$ to $n=2$ is the difference between these two energy levels:

$\Delta E=E_2-E_1$

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

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

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

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

Converting to standard scientific notation to match the options:

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