Q. The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is $(hc\ = 1240\ eV - nm)$
A) 3.09 eV
B) 1.42 eV
C) 151 eV
D) 1.68 eV
Explanation
(b) $\frac{1}{2}\ mv^{2} = eV_{0} = 1.68\ eV$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ hv = \frac{hc}{\lambda} = \frac{1240\ eV - nm\ }{400\ nm} = 3.1 = \ \ eV$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3.1\ eV = \ W_{0} + 1.6eV$
$W_{0} = 1.42eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.