Question #228110

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


1
Expert's answer
2021-08-20T18:04:43-0400

The photoelectric equation is

hcλ=ϕ+EK\frac{hc}{λ} = ϕ+E_K

EK=Kinetic energy

h=plank's constant

c=speed of light

λ=wavelength

ϕ=work function

EK=hcλϕλ=cfrequency=3×1089.0×1014=3.33×107h=6.626×1034  Jsc=3×108  m/sϕ=5.0×1019  JEK=6.626×1034×3×1083.33×1075.0×1019=5.969×10195.0×1019=0.969×1019  J1×1019  JE_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


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