Answer to Question #111054 in Quantum Mechanics for Ananya

Question

threshold for a certain metal is 3600 a when source is placed at 1 m from target. determine the maximum energy in ev of the electron ejected by the radiation of wavelength 2000 a

Answer 1

from photoelectric effect

$\frac{hc}{\lambda}=\frac{hc}{\lambda_0}+KE$

$KE=\frac{1240}{200}-\frac{1240}{360}=6.2eV-3.4eV=2.75eV$

Important conversion used in this question is hc=1240 eV/nm