Q. The wavelength of the energy emitted when electron comes from fourth orbit to second orbit in hydrogen is $20.397cm$. The wavelength of energy for the same transition in $He^{+}$ is
A) $5.099\ cm^{- 1}$
B) $20.497\ cm^{- 1}$
C) $40.994\ cm^{- 1}$
D) $81.988\ cm^{- 1}$
Explanation
(a) $E\left( = \frac{hc}{\lambda} \right) \propto \frac{Z^{2}}{n^{2}} \Rightarrow \lambda \propto \frac{1}{Z^{2}}$
Hence $\lambda_{He^{+}} = \frac{20.397}{4} = 5.099\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.