For a certain metal, when monochromatic light of wavelength is incident, the stopping potential for photoelectrons is

Question

For a certain metal, when monochromatic light of wavelength $λ$ is incident, the stopping potential for photoelectrons is $3 V_{0}$ . When the same metal is illuminated by light of wavelength $2 λ$ , then the stopping potential becomes $V_{0}$ . The threshold wavelength for photoelectric emission for the given metal is $α λ$ . The value of $α$ is _____. [2026]

Smart Achievers · Accepted Answer

(2)

$\frac{h c}{λ} - ϕ = 3 e V_{0}$ $. . . (i)$

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

$From (i) and (ii): \frac{h c}{λ} - ϕ = 3 [\frac{h c}{2 λ} - ϕ]$

$\Rightarrow ϕ = \frac{h c}{4 λ}$

$\frac{h c}{λ_{0}} = \frac{h c}{4 λ} \Rightarrow λ_{0} = 4 λ$

$\Rightarrow α = 4$

Solution

1 1

2 4

3 2

4 3

Ans.
(2)

$\frac{h c}{λ} - ϕ = 3 e V_{0}$ $. . . (i)$

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

$From (i) and (ii): \frac{h c}{λ} - ϕ = 3 [\frac{h c}{2 λ} - ϕ]$

$\Rightarrow ϕ = \frac{h c}{4 λ}$

$\frac{h c}{λ_{0}} = \frac{h c}{4 λ} \Rightarrow λ_{0} = 4 λ$

$\Rightarrow α = 4$

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Solution

1 1

2 4

3 2

4 3

Ans. (2) hcλ-ϕ=3eV0 ...(i) hc2λ-ϕ=eV0 ...(ii) From (i) and (ii): hcλ-ϕ=3[hc2λ-ϕ] ⇒ϕ=hc4λ hcλ0=hc4λ⇒λ0=4λ ⇒α=4

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