When light of wavelength `lambda` is incident on photosensitive surface, the stopping potential is V. When light of wavelength `3lambda` is incident on same surface, the stopping potential is `V/6`Thereshould wave length for the surface is

Options

A. `2lamda`
B. `3lamda`
C. `4lamda`
D. `5lamda`

Answer

D

Explanation

According to quation, using einstein's photoelectric equation `(hc)/(lamda)-phi=eV`
If first case, for `lamda` stopping potential is V
i.e., equation becomes
`(hc)/(lamda)-phi=(eV)/(6)`
`implies (2hc)/(lamda)-phi=eV` . . (ii)
From eqs. (i) and (ii) by subtracting
`(2hc)/(lamda)=eV+6phi implies {:((hc)/(lamda),=eV,+phi),(-," "-,-):}/((hc)/(lamda)=5phi)`
and `phi=(hc)/(lamda_(0))`
thus, `lamda_(0)=5lamda`