Question

Consider a laser pointer that emits red light with wavelength 650 nm. This light is used for a photoelectric effect experiment where the anode in the evacuated glass tube is made up of a material that has work function equal to 1 eV.

[1] What is the energy of an individual photon that comes out of the laser pointer?

[2] What is the maximum kinetic energy of an emitted electron?

Answer 1

1) The energy of a photon is given by the equation E = hc/λ

where E is the energy, Planck's constant h = 6.626 x 10^-34 Js, c = 3.00 x 10⁸ m/s (speed of light), λ is the wavelength of the light (in metres).

So in your case, λ = 650 x 10^-9 m, so the energy of one photon is:

E = (6.626 x 10^-34 3.00 x 10⁸)/650 x 10^-9 = 3.058 10^-19 J = 1.91 eV

2) According to photoelectric effect,
max. K.E of electron = energy of incident photon - work function

max. K.E of electron = 1.91 - 1 = 0.91 eV