A metallic surface is illuminated with radiation of wavelength , the stopping potential is

Question

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-

Smart Achievers · Accepted Answer

(4)

$From the Einstein's photoelectric equation$

$e V_{0} = \frac{h c}{λ} - ϕ_{0} = \frac{h c}{λ} - \frac{h c}{λ_{0}}$ $. . . (i)$

$& \frac{e V_{0}}{4} = \frac{h c}{2 λ} - \frac{h c}{λ_{0}}$ $. . . (i i)$

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

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

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

$\Rightarrow λ_{0} = 3 λ$

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-

1 $\frac{λ}{4}$

2 $4 λ$

3 $\frac{3}{2} λ$

4 $3 λ$

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Solution

Q. A metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V0. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential becomes V04. The threshold wavelength for this metallic surface will be-

1 λ4

2 4λ

3 32λ

4 3λ