Question

The kinetic energy of the most energetic photoelectron emitted from a metal surface is doubled when the wavelength of the incident radiation is reduced from λ₁ to λ₂. The work function of the metal is ______

Options

• `(hc)/(lambda_1lambda_2)(2lambda_2 - lambda_1)`

• `(hc)/(lambda_1lambda_2)(2lambda_1 - lambda_2)`

• `(hc)/(lambda_1lambda_2)(lambda_1 + lambda_2)`

• `(hc)/(lambda_1lambda_2)(lambda_1 - lambda_2)`

Answer 1

The kinetic energy of the most energetic photoelectron emitted from a metal surface is doubled when the wavelength of the incident radiation is reduced from λ₁ to λ₂. The work function of the metal is 

`underline((hc)/(lambda_1lambda_2)(2lambda_2 - lambda_1))`.

Explanation:

Given

K.E._max = `(hc)/lambda_1 - phi` .................(i)

and 2K.E._max = `(hc)/lambda_2 - phi` .................(ii)

Dividing equation (ii) by equation (i), we get

`2 = (((hc)/lambda_2 - phi))/(((hc)/lambda_1 - phi))`

which on solving gives `Φ = (hc)/(lambda_1lambda_2)(2lambda_2 - lambda_1)`.