Question

When a certain metal is illuminated with light of frequency 2.9 ✕ 1015 Hz, a stopping potential of 7.0 V is required to stop the most energetic ejected electrons. What is the work function of this metal?

Answer 1

4.96 eV

Step-by-step explanation:

Recall the Einstein photoeletric equation;

KE = E - Wo

Where;

KE = kinetic energy of the ejected photoelectron

E= energy of the incident photon

Wo = Work function

But

KE = eV= 1.6 * 10^-19 C * 7 V /1.6 * 10^-19 C = 7 eV

E = hf/e = 6.6 * 10^-34 * 2.9 ✕ 10^15 Hz/1.6 * 10^-19

E = 11.96 eV

Hence;

Wo = E - KE

Wo = 11.96 eV - 7eV

Wo = 4.96 eV