Question

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

Answer 1

Given, ionization energy of the neutral $\text{He}$ atom is $24.6\text{eV}$,

From Bohr's atomic model, the ionization energy of a hydrogen like atom is,

$E=13.6\frac{Z^2}{n^2}.......\left(1\right)$

From a neutral $\text{He}$ atom, energy required to remove the first electron is $=24.6\text{eV}$

Now it becomes a single electron atom $\left({\text{He}}^{+}\right)$ and we can apply Bohr's model.

Using (1) we get,

$E=13.6\frac{(2{)}^2}{(1{)}^2}$

$\Rightarrow E=13.6\times 4=54.4\text{eV}$

$\text{Total energy required}=24.6+54.4=79\text{eV}$