Nuclear & Particle Physics - Photoelectric

Light of wavelength 500 nanometers is incident on sodium, with work function 2.28 electron volts. What is the maximum kinetic energy of the ejected photoelectrons?

A. 0.03 eV
B. 0.2 eV
C. 0.6 eV
D. 1.3 eV
E. 2.0 eV

(GR9677 #42)

Solution:

Einstein’s photoelectric equation:  |eV| = hv − W = ^hc/_λ − W

h = 4.1 × 10⁻¹⁵ eV.second
c = 3 × 10⁸ m/s
λ = 500 nm = 5 × 10⁻⁷ m
W = 2.28 eV

KE = [(4.1 × 10⁻¹⁵)(3 × 10⁸)/(5 × 10⁻⁷ )] − 2.28 
=  [(4.1 × 3 / 5)(10⁻¹⁵ × 10⁸ × 10⁷)] − 2.28
=  2.46 − 2.28 = 0.18 ≈  0.2 eV

Answer: B