Photoelectrons from sodium at 345 nm

Problem

Light of wavelength $345\;\text{nm}$ falls on a sodium surface ($W_0 = 2{,}28\;\text{eV}$). Calculate the maximum kinetic energy of the emitted photoelectrons.

Solution

Step 1 — Photon energy.

$E_f = \frac{hc}{\lambda} = \frac{1240\;\text{eV·nm}}{345\;\text{nm}} \approx 3{,}59\;\text{eV}$

Step 2 — Einstein’s formula. The maximum kinetic energy is the excess over the work function:

$E_K = E_f - W_0 = 3{,}59 - 2{,}28$

$\ev{E_K \approx 1{,}30\;\text{eV}}$

Since $E_f = 3{,}59\;\text{eV} > W_0 = 2{,}28\;\text{eV}$, emission does occur: the electrons come out with at most $1{,}30\;\text{eV}$ of kinetic energy.

Links

Topics: Quantum physics Concepts: Photoelectric effect · Photon