In a certain experiment, the photoelectric cut-off voltage is 3.5 V.  What will be the maximum kinetic energy of the photoelectrons emitted?

5.4 × 10-19 J 0.54 × 10-19 J 3.1 × 10-19 J 5.4 × 10-20 J

5.4 × 10-19 J

The maximum kinetic energy of the photoelectrons is given by Kmax = eV0 V0     = 3.5 V e = 1.6 × 10-19 C Kmax = 3.5 × 1.6 × 10-19 J         =5.4 × 10-19 J