When a metallic surface is illuminated with radiation of wavelength , the stopping potential is . If the same surface is illuminated with radiation of wavelength , the stopping potential is . The threshold wavelength for the metallic surface is

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Solution

Q.
When a metallic surface is illuminated with radiation of wavelength $λ$ , the stopping potential is $V$ . If the same surface is illuminated with radiation of wavelength $2 λ$ , the stopping potential is $\frac{V}{4}$ . The threshold wavelength for the metallic surface is [2016]

1 $\frac{5}{2} λ$

2 $3 λ$

3 $4 λ$

4 $5 λ$

Ans.
(2)

According to Einstein's photoelectric equation,

$e V_{s} = \frac{h c}{λ} - \frac{h c}{λ_{0}}$

$∴$ $As per question, e V = \frac{h c}{λ} - \frac{h c}{λ_{0}}$ ...(i)

$\frac{e V}{4} = \frac{h c}{2 λ} - \frac{h c}{λ_{0}}$ ...(ii)

From equations (i) and (ii), we get

$\frac{h c}{2 λ} - \frac{h c}{4 λ} = \frac{h c}{λ_{0}} - \frac{h c}{4 λ_{0}} \Rightarrow \frac{h c}{4 λ} = \frac{3 h c}{4 λ_{0}}$ or $λ_{0} = 3 λ$

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