Question

Consider a beam of light with a wavelength λ = 403-nm incident onto a metal surface, which can be Li, Be or Hg. The work functions of these metals are 2.30-eV, 3.90-eV and 4.50-eV respectively. For the metal that exhibits the photoelectric effect find the maximum kinetic energy of the photoelectrons.

Answer 1

max KE = hc/L- Wo

Where h is plancks constant = 6.626e -34 Js

c is speed of light = 3e8 m/s

L is wavelength

so
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for Wo = 2.3 eV

KEmax = (6.626 e -34 * 3e8)/(403 e-9) - 2.3 eV

KEmax = (4.93 e 19/1.6 e -19) eV - 2.3 eV

KEmax = 3.08 eV - 2.3 eV = 0.781 eV -----<<<<<<<<<<Answer

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for Wo = 3.9 eV

KEmax = 3.08 eV - 3.9 eV = -0.82 eV -----<<<<<<<<<<Answer

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for Wo = 4.5 eV

KEmax = 3.08eV - 4.5 eV = - 1.42 eV -----<<<<<<<<<<Answer

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