In a certain experiment, the photoelectric cut-off voltage is 3.5 V. What will be the maximum kinetic energy of the photoelectrons emitted?
Answer & explanation
Correct answer: option 1
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