Question

A light whose frequency is equal to 6 x 104 Hz is incident on a metal whose work function is 2 eV.
The maximum energy of electron emitted is :
(A) 2.49 eV
(B) 4.49 eV
(C) 0.49 eV
(D) 5.49 eV​

Answer 1

Answer:

Absorbed energy = Threshold energy + Kinetic energy of photoelectrons

Absorbed energy =hv

=6.626×10

−34

×6×10

14

=3.9756×10

−19

J

=

1.6×10

−19

3.9756×10

−19

=2.49eV

2.49=2eV+ Kinetic energy of photoelectron

Kinetic energy of photoelectron =0.49eV

Explanation:

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