The wavelength of the radiation emitted with an electron jumps from the fourth orbit to the second orbit in an hydrogen atom is 20.36 cm. what is the wavelength of radiation emitted for the same transition in `He^+` ?

Options

A. 10.18 cm
B. 40.72 cm
C. 5.09 cm
D. 81.44 cm

Answer

C

Explanation

The wavelength fo emitted radiation is given by
`1/lamda = RZ^2 ((1)/(n_i^2) - (1)/(n_f^2))`
Thus `1/lamda prop Z^2 " or " lamda prop 1/Z^2`
For helium, Z=2 `therefore lamda_(He) = (20.36)/(4) = 5.09 cm`