A certain metal having a work function of 5.4 eV is illuminated with light of frequency 2.7 x 1015Hz. Determine the stopping potential (in V) that is required to stop the most energetic ejected electrons. Your answer should be a number with two decimal places, do not include unit.

5.79 Volts

Explanation:

W = work function of the metal = 5.4 eV = 5.4 x 1.6 x 10⁻¹⁹ J = 8.64 x 10⁻¹⁹ J

f = frequency of light = 2.7 x 10¹⁵ Hz

V₀ = stopping potential for the electron = ?

q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

Using conservation of energy

e V₀ = hf - W

(1.6 x 10⁻¹⁹) V₀ = (6.63 x 10⁻³⁴) (2.7 x 10¹⁵) - 8.64 x 10⁻¹⁹

V₀ = 5.79 Volts