When a surface 1 cm thick is illuminated with light of wavelength `lambda`, the stopping potential is `V_(0)`, but when the same surface is illuminated by light of wavelength `3lambda`, the stopping potential is `(V_(0))/(6)`. Find the threshold wavelength for metallic surface.

Options

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

Answer

B

Explanation

From Einstein's photoelectric equation, we have
`eV_(0)=hc[(1)/(lamda)-(1)/(lamda_(0))]` . . . (i)
`(eV_(0))/(6)=hc[(1)/(3lamda)-(1)/(lamda_(0))]` . . (ii)
Dividing eq. (i) by Eq. (ii) we get
`6=(((1)/(lamda)-(1)/(lamda_(0))))/(((1)/(3lamda)-(1)/(lamda_(0))))implies (6)/(3lamda)-(6)/(lamda_(0))=(1)/(lamda)-(1)/(lamda_(0))`
or `(1)/(lamda)=(5)/(lamda_(0))implies (lamda_(0))/(lamda)=5 implies lamda_(0)=5lamda`