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The light intensity incident on a metallic surface with a work function of 1.88 eV produces photoelectrons with a maximum kinetic energy of
Question
The light intensity incident on a metallic surface with a work function of 1.88 eV produces photoelectrons with a maximum kinetic energy of 1.2 eV.
if the frequency of the light is doubled, what is the maximum kinetic energy of the photoelectrons?
Answer in units of eV.
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Physics
3 years
2021-08-09T14:08:39+00:00
2021-08-09T14:08:39+00:00 1 Answers
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Answers ( )
Answer:
KE = KE (incidental) – KE of emitted photons
or KE = h * f – Wf
So h * f = KE + Wf = 1.2 + 1.88 = 3.08 incident energy
If you double the frequency then h * f = 6.16
KE = 6.16 – 1.2 = 4.96 eV