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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Khoii Minh 3 years 2021-08-09T14:08:39+00:00 1 Answers 40 views 1

Answers ( )

  1. mit
    0
    2021-08-09T14:09:56+00:00
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    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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