Photoelectric cut-off voltage is given as $\mathbf{V}_{\mathbf{0}}=\mathbf{1 . 5} \mathbf{V}$

The maximum kinetic energy for emitted photoelectrons,

$K_{e}=eV$

Where,

$e=$ charge on an electron $=1.6 \times 10^{-19} \mathrm{C}$

Therefore, $K_{e}=1.6 \times 10^{-19} \times 1.5=2.4 \times 10^{-19} \mathrm{~J}$

Therefore, the maximum kinetic energy emitted by the photoelectrons is $2.4 \times 10^{-19} \mathrm{~J}$