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 $47.2\, eV$, 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
$E_{n}=\frac{-R h C z^{2}}{n^{2}}$
For hydrogen atom, $z=1$ and $n=1$, hence
$E_{1}=\frac{-R h C \cdot 1^{2}}{1^{2}}=-R h C$
We know that, $R h C=13.6\, eV$
$\therefore E_{1}=-13.6\, eV$
For $n=2$.
$E_{2}=\frac{-R h C z^{2}}{2^{2}}=\frac{-136 z^{2}}{4}$
$\Rightarrow E_{2}=\frac{-13.6 z^{2}}{4} ...(i)$
Similarly, for $n=3$,
$E_{3}=\frac{-13.6}{9} z^{2} ...(ii)$
Given $\Delta E=E_{3}-E_{2}=47.2$
$\Rightarrow \frac{-13.6}{9} z^{2}-\left(\frac{-13.6 z^{2}}{4}\right)=47.2$
$\Rightarrow 13.6 \times \frac{5}{36} z^{2}=47.2$
$\Rightarrow z^{2}=24.98$
$\Rightarrow z^{2} \simeq 25$
$\Rightarrow z=5$