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A certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for photo-electric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$, the stopping potential is $V_0$. The threshold wavelength for this surface for photo-electric effect is
A
$6\lambda$
B
$4\lambda$
C
$\frac{\lambda}{4}$
D
$\frac{\lambda}{6}$
Detailed Solution
$eV_0 = \frac{hc}{\lambda} - W$
For $\lambda$: $3eV_0 = \frac{hc}{\lambda} - W$ ...(1)
For $2\lambda$: $eV_0 = \frac{hc}{2\lambda} - W$ ...(2)
Substituting (2) in (1): $\frac{3hc}{2\lambda} - 3W = \frac{hc}{\lambda} - W \Rightarrow \frac{hc}{2\lambda} = 2W \Rightarrow W = \frac{hc}{4\lambda}$
Threshold wavelength $\lambda_0 = \frac{hc}{W} = 4\lambda$
For $\lambda$: $3eV_0 = \frac{hc}{\lambda} - W$ ...(1)
For $2\lambda$: $eV_0 = \frac{hc}{2\lambda} - W$ ...(2)
Substituting (2) in (1): $\frac{3hc}{2\lambda} - 3W = \frac{hc}{\lambda} - W \Rightarrow \frac{hc}{2\lambda} = 2W \Rightarrow W = \frac{hc}{4\lambda}$
Threshold wavelength $\lambda_0 = \frac{hc}{W} = 4\lambda$