Answer to Question #75369 in Quantum Mechanics for Deepesh Tripathi

Question

A photon of energy 8ev is incident on a metal surface of threshold frequency 1.6/1000000000000000 Hz . The kinetic energy of the photoelectrons emitted is (in ev ) (1) 6. (2) 1.6 (3) 1.2. (4) 2

Answer 1

Minimum frequency (red limit) from which substance the photoelectric effect begins:
ν_min=W/h
We find the work of the W output, that is, the work that must be performed to remove the electron from the surface of the metal, and the transmission of the kinetic energy electron:
W=〖hν〗_min
W=(6×〖10〗^(-34) Js×1.6×〖10〗^15 s^(-1))/(1.6×〖10〗^(-19) J)=6 eV
The kinetic energy of a photoelectron can be found using the Einstein equation for a photoelectric effect
hν=W+KE
We get
KE=hν-W
Finally
KE=8 eV-6 eV=2 eV