Physics, asked by noorjotsinghman5828, 10 months ago

Light of energy 4.7 1019 j is irradiated on a metal surface having work function equal to 2.02 ev. The kinetic energy of the emitted photoelectrons is

Answers

Answered by hannjr
1

Answer:

1 eV = 1.6 * 10E-19 J      (charge of electron * 1 V)

2.02 eV = 2.02 * 1.6 * 10E-19 J = 3.23 J

h f = 4.7 * 10E-19 J        energy of incident light

(4.7 - 3.23) * 10E-19 J = 1.5 * 10E-19 J    the KE of the phoelectrons

where KE = h f - Wf     where Wf is the work function

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