The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

Photoelectric cut-off voltage, Vo = 1.5 V

The maximum kinetic energy of the emitted photoelectrons is given as:

K_e=eV₀

Where, e = Charge on an electron = 1.6 x 10^-19 C

Therefore, K_e = 1.6 x 10^-19 x 1.5

= 2.4 x 10^-19 J

Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 x 10^-19J.