The photoelectric cut-off voltage in a certain experiment is 1.5 V.

Photoelectric cut-off voltage, V0 = 1.5 V

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

$K_{e}=e V_{0}$

Where,

e = Charge on an electron = 1.6 × 10−19 C

$\therefore K_{e}=1.6 \times 10^{-19} \times 1.5$

$=2.4 \times 10^{-19} \mathrm{~J}$

Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 × 10−19 J.