The wavelength of the energy emitted when electron come from fourth orbit to second orbit in hydrogen is 20.397 cm. The wavelength of energy for the same transition in He+ is
Answer & explanation
Correct answer: option 1
$E\left(=\frac{h c}{\lambda}\right) \propto \frac{Z^2}{n^2} \Rightarrow \lambda \propto \frac{1}{Z^2}$
Hence $\lambda_{He^{+}}=\frac{20.397}{4}$ = 5.099 cm