Physic Labs

Problem 3

Monochromatic light of wavelength λ=400\lambda=400 nm illuminates a metal surface with work function W=2.00W=2.00 eV. Use hc=1240hc=1240 eV nm and electron charge magnitude ee. Assume sufficient intensity and photoelectrons emitted in all directions. (a) Calculate the photon energy and maximum photoelectron kinetic energy. (b) Find the stopping potential needed so no photoelectron reaches the collector. (c) For the same metal, calculate the longest wavelength that still produces photoelectrons. (d) Change the light to λ=300\lambda=300 nm and find the new maximum kinetic energy.
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Eγ=hcλ=1240 eV nm400 nm=3.10 eVE_\gamma=\frac{hc}{\lambda}=\frac{1240\ \mathrm{eV\,nm}}{400\ \mathrm{nm}}=3.10\ \mathrm{eV}
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Each photon carries energy hc/λhc/\lambda. Since hchc is supplied in eV nm, dividing by λ\lambda in nm returns the energy directly in eV.