When a centimeter thick surface is illuminated with light of wavelength `lamda`, the stopping potential is V. When the same surface is illuminated by light of wavelength `2lamda`, the stopping potential is `(V)/(3)`. Threshold wavelength for the metallic surface is

Options

A. `4lamda//3`
B. `4lamda`
C. `6lamda`
D. `lamda//3`

Answer

B

Explanation

According to the question,
`eV=hc((1)/(lamda)-(1)/(lamda_(0)))` . .. (i)
and `(eV)/(3)= hc ((1)/(2lamda)-(1)/(lamada_(0)))` . .. (ii)
Dividing Eq. (i) and Eq. (ii), we get
`3=(((1)/(lamda)-(1)/(lamda_(0))))/(((1)/(2lamda)-(1)/(lamda_(0)))) and 3 ((1)/(2lamda)-(1)/(lamda_(0)))=(1)/(lamda)-(1)/(lamda_(0))`
or `(3)/(2lamda)-(1)/(lamda)=(3)/(lamda_(0))-(1)/(lamda_(0)) implies (1)/(2lamda)=(2)/(lamda_(0))`
threshold wavelength for metallic surface `lamda_(0)=4lamda`.