An energy of $24.6\text{eV}$ is required to remove one of the electron from the ground state of a neutral helium atom. The energy $\left(\text{in eV}\right)$ required to remove both the electrons from the ground state of a neutral helium atom is:

The correct option is A $79.0\text{eV}$
The energy required to remove the first electron is given as, $E=24.6\text{eV}$

After the removal of first electron, Helium becomes hydrogen like atom $\left(H{e}^{+}\right)$.
$\therefore Z=2$ and $n=1$

So the energy of the remaining single electron is given by,

$E=\left(\frac{Z^2}{n^2}\right)(-13.6\text{eV})=\frac{-13.6\text{eV}\times 4}{1}=-54.4\text{eV}$

Negative sign indicates that the electron is bound to the nucleus.

Hence, an energy of $E_1=54.4\text{eV}$ is required to remove the second electron from the ground state of the neutral helium atom.

So, the total energy required,

$E=24.6\text{eV}+54.4\text{eV}=79\text{eV}$

Hence, option $\left(A\right)$ is correct.

Why this Question? Note: 1. The energy of electrons revolving around the nucleus is taken as negative because it is bounded due to attractive forces between electron and nucleus. 2. The attractive forces are always taken as negative while the repulsive forces are taken as positive. The energy required to pull an electron out of the atom is positive.