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When a metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is V. If the same surface is illuminated with radiation of wavelength $2\lambda$, the stopping potential is $\frac{V}{4}$. The threshold wavelength for the metallic surface is
A
$5\lambda$
B
$\frac{5}{2}\lambda$
C
$3\lambda$
D
$4\lambda$
Explanation
Write Einstein's equation for both wavelengths and eliminate eV.
Detailed Solution
Photoelectric equation: $eV = \frac{hc}{\lambda} - \phi$, where $\phi = \frac{hc}{\lambda_{th}}$
Case I: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_{th}}$ ...(i)
Case II: $\frac{eV}{4} = \frac{hc}{2\lambda} - \frac{hc}{\lambda_{th}} \Rightarrow eV = \frac{2hc}{\lambda} - \frac{4hc}{\lambda_{th}}$ ...(ii)
Subtracting (ii) from (i): $0 = -\frac{hc}{\lambda} + \frac{3hc}{\lambda_{th}} \Rightarrow \frac{3hc}{\lambda_{th}} = \frac{hc}{\lambda}$
$\lambda_{th} = 3\lambda$
Case I: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_{th}}$ ...(i)
Case II: $\frac{eV}{4} = \frac{hc}{2\lambda} - \frac{hc}{\lambda_{th}} \Rightarrow eV = \frac{2hc}{\lambda} - \frac{4hc}{\lambda_{th}}$ ...(ii)
Subtracting (ii) from (i): $0 = -\frac{hc}{\lambda} + \frac{3hc}{\lambda_{th}} \Rightarrow \frac{3hc}{\lambda_{th}} = \frac{hc}{\lambda}$
$\lambda_{th} = 3\lambda$