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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
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