Question

The wavelength of the energy emitted when an electron comes from the fourth orbit to the second orbit in hydrogen is $20.397cm.$ The wavelength of energy for the same transition in $H{e}^{+}$ is

• A. $5.099c{m}^{-1}$
• B. $20.497c{m}^{-1}$
• C. $40.994c{m}^{-1}$
• D. $81.988c{m}^{-1}$

Answer 1

The correct option is A $5.099c{m}^{-1}$

$E(=\frac{hc}{\lambda })\propto \frac{Z^2}{n^2}$

$\Rightarrow \lambda \propto \frac{1}{Z^2}$

Hence ${\lambda }_{H{e}^{+}}=\frac{20.397}{4}=5.099c{m}^{-1}$