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3 years ago

6

The light intensity incident on a metallic surface with a work function of 1.88 eV produces photoelectrons with a maximum kineti

c 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.

1 answer:

ivanzaharov [21]3 years ago

7 0

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

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