A photon of energy 10.2 eV corresponds to light of wavelength `lamda_(0)`. Due to an electron transition from n=2 to n=1 in a hydrogen atom, light of wavelength `lamda` is emitted. If we take into account the recoil of the atom when the photon is emitted.

Options

A. `lamda=lamda_(0)`
B. `lamda lt lamda_(0)`
C. `lamda gt lamda_(0)`
D. the data is not sufficient to reach a conclusion

Answer

C

Explanation

Without recoil of atom, `(hc)/(lamda_(0))=10.2`
With recoil :
`(hc)/(lamda)+(1)/(2) mv^(2)=10.2`
`(hc)/(lamda) lt (hc)/(lamda_(0))`
`lamda_(0) lt lamda implies lamda gt lamda_(0)`