Q. Figure represents a graph of photo current $I$ versus applied voltage $(V).$ The maximum energy of the emitted photoelectrons is
A) $2eV$
B) $4\ eV$
C) $0\ eV$
D) $4\ J$
Explanation
(b) From the graph stopping potential $\left| V_{s} \right| = - V$
Also $k_{\max} = \left( \left| V_{0} \right| \right)eV = 4eV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.