The energy required to excite an electron from $n=2$ to $n=3$ energy state is $47.2\text{eV}$. The charge number of the nucleus, around which the electron revolving, will be-

The correct option is D $5$
The expression of total energy for hydrogen like atom is,

$E_n=-13.6\frac{Z^2}{n^2}$

$\Rightarrow \Delta E=E_3-E_2=13.6Z^2[\frac{1}{n_2^2}-\frac{1}{n_3^2}]$

$47.2=13.6Z^2[\frac{1}{2^2}-\frac{1}{3^2}]$

$47.2=13.6Z^2\left[\frac{9-4}{4\times 9}\right]$

$Z^2=\frac{472}{136}\times \frac{36}{5}=\frac{118\times 36}{34\times 5}$

$\Rightarrow Z^2=24.988\approx 25$

i.e., $5$ protons are present in the nucleus.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.