When a metal surface is illuminated by light of wavelength , the stopping potential is 8V. When the same surface is illuminated by light of wavelength , stopping potential is 2V. The threshold wavelength for this surface is

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Solution

Q.
When a metal surface is illuminated by light of wavelength $λ$ , the stopping potential is 8V. When the same surface is illuminated by light of wavelength $3 λ$ , stopping potential is 2V. The threshold wavelength for this surface is [2024]

1 $9 λ$

2 $3 λ$

3 $4.5 λ$

4 $5 λ$

Ans.
(1)

$E = ϕ + K_{\max}$

$ϕ = \frac{h c}{λ_{0}}$

$K_{\max} = e V_{0}$

$8 e = \frac{h c}{λ} - \frac{h c}{λ_{0}}$ ...(i)

$2 e = \frac{h c}{3 λ} - \frac{h c}{λ_{0}}$ ...(ii)

On solving (i) and (ii) $λ_{0} = 9 λ$

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