Question

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

• A. $5.099\mspace{6mu} cm^{- 1}$
• B. $20.497\mspace{6mu} cm^{- 1}$
• C. $40.994\mspace{6mu} cm^{- 1}$
• D. $81.988\mspace{6mu} cm^{- 1}$

Answer 1

Answer: (A)

$5.099\mspace{6mu} cm^{- 1}$

Answer 2

$E\left( = \frac{hc}{\lambda} \right) \propto \frac{Z^{2}}{n^{2}}$ ⇒ $\lambda \propto \frac{1}{Z^{2}}$

Hence $\lambda_{He^{+}} = \frac{20.397}{4} = 5.099cm$