Question

Photoelectric emission is observed from a metallic surface for frequencies ν₁ and ν₂ of the incident light (ν₁ > ν₂). If the maximum value of kinetic energy of the photoelectrons emitted in the two cases are in the ration 1 : n then the threshold frequency of the metallic surface is ______.

Options

• `(nu_1 - nu_2)/("n - 1")`

• `("n "nu_1 - nu_2)/("n - 1")`

• `("n " nu_2 - nu_1)/("n - 1")`

• `(nu_1 - nu_2)/("n")`

Answer 1

Photoelectric emission is observed from a metallic surface for frequencies ν₁ and ν₂ of the incident light (ν₁ > ν₂). If the maximum value of kinetic energy of the photoelectrons emitted in the two cases are in the ration 1 : n ·then the threshold frequency of the metallic surface is `underline(("n "nu_1 - nu_2)/("n - 1"))`.

Explanation:

Photoelectric equation hν - hν₀ = K_max

h(ν₁ - ν₀) = K₁    ....(1) and h(ν₂ - ν₀) = K  ...(2)

from eq (1) and (2) we get

`(ν_1 - ν_0)/(ν_2 - ν_0) = 1/"n" => "n"nu_1 - "n"nu_0 = "v"_2 - "v"_0`

`(ν_1 - ν_0)/(ν_2 - ν_0) = 1/"n" => "n"ν_1 - ν_2 = "n"ν_0 - "v"_0`

`(("n"ν_1 - ν_2))/("n - 1") = ν_0`