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Stopping Potential

When a certai...

Question

When a certain metal surface is illuminated with light of frequency $ν,$ the stopping potential for photoelectric current is $V_{0}$ . When the same surface is illuminated by light of frequency $\frac{ν}{2}$ the stopping potential is $\frac{V_{0}}{4}$ . The threshold frequency for photoelectric emission is :

ν / 6

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ν / 3

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2 ν / 3

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4 ν / 3

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Solution

The correct option is B $ν / 3$
Incident frequency $= ν$

Stopping potential $= V_{0}$

$∴$ Using Photoelectric effect :

$E = ϕ + (K E)_{m a x}$

$\Rightarrow h ν = ϕ + e V_{0}$ ------ $(1)$

and similarly ,

$\frac{h ν}{2} = ϕ + \frac{e V_{0}}{4}$ ------- $(2)$

from $(2)$

$h ν = 2 ϕ + \frac{e V_{0}}{2}$ ----- $(3)$

from $(1)$ and $(3)$

$h ν = 2 ϕ + \frac{1}{2} [h ν - ϕ]$

$\Rightarrow h ν = 2 ϕ + \frac{h ν}{2} - \frac{ϕ}{2}$

$= (2 ϕ - \frac{ϕ}{2}) + \frac{h ν}{2}$

$\Rightarrow \frac{h ν}{2} = \frac{3}{2} ϕ$

$ϕ = \frac{h ν}{3}$

$∴$ frequency $= \frac{ν}{3}$

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