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 becomes . The threshold wavelength for this metallic surface will be

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Solution

Q.
A metallic surface is illuminated with radiation of wavelength $λ$ , the stopping potential is $V_{0}$ . If the same surface is illuminated with radiation of wavelength $2 λ$ , the stopping potential becomes $\frac{V_{0}}{4}$ . The threshold wavelength for this metallic surface will be [2023]

1 $\frac{λ}{4}$

2 $4 λ$

3 $\frac{3}{2} λ$

4 $3 λ$

Ans.
(4)

$From the equation of photoelectric effect$

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

$\frac{e V_{0}}{4} = \frac{h c}{2 λ} - \frac{h c}{λ_{0}}$

$\Rightarrow \frac{1}{4} (\frac{h c}{λ} - \frac{h c}{λ_{0}}) = \frac{h c}{2 λ} - \frac{h c}{λ_{0}}$

$\frac{1}{λ_{0}} - \frac{1}{4 λ_{0}} = \frac{1}{2 λ} - \frac{1}{4 λ}$

$\frac{3}{4 λ_{0}} = \frac{1}{4 λ} \Rightarrow λ_{0} = 3 λ$

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