Question

A hydrogen like atom has one electron revolving round a stationary nucleus. If the energy required to excite the electron from the $2^{\text{nd }}$ orbit to $3^{\text{rd }}$ orbit is $47.2eV$, find the atomic number of the given atom

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (C)

The energy of $n$th orbit is given as
$E_n=\frac{-RhC{z}^2}{n^2}$
For hydrogen atom, $z=1$ and $n=1$, hence
$E_1=\frac{-RhC\cdot 1^2}{1^2}=-RhC$
We know that, $RhC=13.6eV$
$\therefore E_1=-13.6eV$
For $n=2$.
$E_2=\frac{-RhC{z}^2}{2^2}=\frac{-136z^2}{4}$
$\Rightarrow E_2=\frac{-13.6z^2}{4}...(i)$
Similarly, for $n=3$,
$E_3=\frac{-13.6}{9}{z}^2...(ii)$
Given $\Delta E=E_3-E_2=47.2$
$\Rightarrow \frac{-13.6}{9}{z}^2-\left(\frac{-13.6z^2}{4}\right)=47.2$
$\Rightarrow 13.6\times \frac{5}{36}{z}^2=47.2$
$\Rightarrow z^2=24.98$
$\Rightarrow z^2\simeq 25$
$\Rightarrow z=5$