Question

When a photosensitive surface is irradiated by lights of wavelengths `lambda_1` and `lambda_2`, kinetic energies of emitted photoelectrons are E₁ and E₂ respectively. The work function of the photosensitive surface is ____________.

Options

• `(lambda_2 "E"_1 + lambda_2 "E"_2)/(lambda_2 - lambda_1)`

• `(lambda_2 "E"_2 + lambda_1 "E"_1)/(lambda_2 - lambda_1)`

• `(lambda_1 "E"_1 + lambda_2 "E"_2)/(lambda_2 - lambda_1)`

• `(lambda_1 "E"_1 + lambda_2 "E"_2)/(lambda_2 - lambda_1)`

Answer 1

When a photosensitive surface is irradiated by lights of wavelengths `lambda_1` and `lambda_2`, kinetic energies of emitted photoelectrons are E₁ and E₂ respectively. The work function of the photosensitive surface is `(lambda_1 "E"_1 + lambda_2 "E"_2)/(lambda_2 - lambda_1)`.

Explanation:

We know,

`"hv"_1 = "E"_1 + omega_0`

`therefore "hc"/lambda_1 = "E"_1 + omega_0`

`"hc" = ("E"_1 + omega_0) lambda_1`    ....(1)

`"Similarly, hc" = ("E"_2 + omega_0)lambda_2`    ....(2)

`therefore omega_0 lambda_1 + "E"_1 lambda_1 = omega_0 lambda_2 + "E"_2 lambda_2`

`omega_0 (lambda_1 - lambda_2) = "E"_2 lambda_2 - "E"_1 lambda_1`

`omega_0 = ("E"_2 lambda_2 - "E"_1 lambda_1)/(lambda_1 - lambda_2)`

`= ("E"_1 lambda_1 - "E"_2 lambda_2)/(lambda_2 - lambda_1)`