Threshold wavelength of a metal is `lamda_(0)`. The de Broglie wavelength of photoelectron when the metal is irradiated with the radiation of wavelength `lamda` is:

Options

A. `[(h lamdalamda_(0))/(2cm)]^((1)/(2))`
B. `[(h(lamda-lamda_(0)))/(2cmlamdalamda_(0))]^(1//2)`
C. `[(h(lamda_(0)-lamda))/(2cmlamdalamda_(0))]^(1//2)`
D. `[(hlamdalamda_(0))/(2mc(lamda_(0)-lamda))]^(1//2)`

Answer

D

Explanation

Absorbed energy=Threshold energy+Kinetic energy of photoelectron
`h(c)/(lamda)=(hc)/(lamda_(0))+E`
`E=hc[(1)/(lamda)-(1)/(lamda_(0))]` . . . .(i)
de Broglie wavelength can be calculated as
`lamda=(h)/(sqrt(2Em))` .. . . (ii)
From eqs. (i) and (ii)
`lamda=[(hlamdalamda_(0))/(2mc(lamda_(0)-lamda))]^(1//2)`