Answer to Question #228110 in Atomic and Nuclear Physics for Mis

Question

Electromagnetic radiation of frequency 9.0 x 10^14Hz is incident on a clean metal surface. The work function of the metal is 5.0 x 10^-19J. Determine the maximum kinetic energy of photoelectrons released from the metal surface

Answer 1

The photoelectric equation is

$\frac{hc}{λ} = ϕ+E_K$

EK=Kinetic energy

h=plank's constant

c=speed of light

λ=wavelength

ϕ=work function

$E_K = \frac{hc}{λ} -ϕ λ = \frac{c}{frequency} \\ = \frac{3 \times 10^8}{9.0 \times 10^{14}} \\ = 3.33 \times 10^{-7} \\ h=6.626 \times 10^{-34} \;Js \\ c= 3 \times 10^8 \;m/s \\ ϕ = 5.0 \times 10^{-19} \;J \\ E_K = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.33 \times 10^{-7}} -5.0 \times 10^{-19} \\ = 5.969 \times 10^{-19} -5.0 \times 10^{-19} \\ = 0.969 \times 10^{-19} \;J \\ ≈ 1 \times 10^{-19} \;J$