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

Options

A. 5.099 cm
B. 20.497 cm
C. 40.994 cm
D. 81.988 cm

Answer

A

Explanation

`E= (hc)/(lamda) prop (Z^(2))/(n^(2)) rArr lamda prop (1)/(Z^(2))`
`rArr (lamda_(He))/(lamda_(H)) = (Z_(H)^(2))/(Z_(He)^(2)) rArr lamda_(He) = (20.397)/(4)= 5.099cm`