Question

18. A light whose frequency is equal to 6 x 1014 Hz is incident on a metal whose work function is 2eV
(h=6.63 x 10-34 Js, leV=1.6 10-19 J). The maximum energy of electrons emitted will be:
A) 2.49 eV B) 4.49 eV C) 0.49 eV
D) 5.49 eV
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Answer 1

Answer:

Explanation:

Absorbed energy = Threshold energy + Kinetic energy of photoelectrons

Absorbed energy =hv

 =6.626×10  −34  ×6×10  14

=3.9756×10  −19  

=  1.6×10  −19

 3.9756×10 −19

 =2.49eV

2.49=2eV+ Kinetic energy of photoelectron

Kinetic energy of photoelectron =0.49eV