When a certain metallic surface is illuminated with monochromatic light of wavelength `lamda`, the stopping potential for photoelectric current is `3V_0` and when the same surface is illuminated with light of wavelength `2lamda`, the stopping potential is `V_0`. The threshold wavelength of this surface for photoelectrice effect is

Options

A. `6lamda`
B. `(4lamda)/(3)`
C. `4lamda`
D. `8lamda`

Answer

C

Explanation

Using photoelectric equation,
`hv-hv_0=(1)/(2)mv^2=eV_s`
or `((hc)/(lamda)-(hc)/(lamda_0))=eV_s`
For the first case, `(hc)/(2lamda)-(hc)/(lamda_0)=e(3V_0)` ...(i)
For the second case, `(hc)/(2lamda)-(hc)/(lamda_0)=e(eV_0)`..(ii)
Solving `lamda_0=4lamda`