Q. The wave number of the energy emitted when electron comes from fourth orbit to second orbit in hydrogen is $20,397\ cm^{- 1}$. The wave number of the energy for the same transition in $He^{+}$is
A) $5,099\ cm^{- 1}$
B) $20,497\ cm^{- 1}$
C) $14400\ \mathring{\mathrm{A}}$
D) $81,588\ cm^{- 1}$
Explanation
(d) $\overline{v} = R\left\lbrack \frac{1}{2^{2}} - \frac{1}{4^{2}} \right\rbrack = \frac{3R}{4} = 20397cm^{- 1}$
For the same transaction in He atom $(Z = 2)$
$\overline{v} = RZ^{2}\left\lbrack \frac{1}{2^{2}} - \frac{1}{4^{2}} \right\rbrack = \frac{3R \times 2^{2}}{4}$
$= 20397 \times 4 = 81588\ cm^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.