Question

Radiation of wavelength λ is incident on a photosensitive surface. Find the de Broglie wavelength of electrons emitted from the surface. Assume that the work function of the surface is negligible.

Answer 1

The incident photon transfers its energy to the electron. The kinetic energy of the emitted photoelectron determines its de Broglie wavelength.

Since the work function Φ is negligible (Φ ≈ 0):

The maximum kinetic energy of the emitted electron is equal to the energy of the incident photon:

K = `(h c)/lambda`

Now, substituting this value of kinetic energy into the de Broglie wavelength formula:

λ_e = `h/(sqrt(2 m K)`

= `h/(sqrt(2 m((h c)/lambda))`

= `(h * sqrt lambda)/(sqrt(2m h c))`

= `sqrt((h^2 lambda)/(2m h c))`

= `sqrt((h lambda)/(2m c))`

∴ The de Broglie wavelength of the emitted electrons is `sqrt((h lambda)/(2m c))`.