Question 11.3: The photoelectric cut-off voltage in a certain experiment is 1.5 V.
What is the maximum kinetic energy of photoelectrons emitted?

Hint: $$K.E._{max}=eV_{0}$$

Step: Find the maximum kinetic energy of photoelectrons emitted.
Photoelectric cut-off voltage, V0 = 1.5 V
For emitted photoelectrons, the maximum kinetic energy is:
$$K.E._{max}=eV_{0}$$
$$\Rightarrow~K.E._{max}=eV_{0}=1.6\times10^{-19}\times1.5$$
$$\Rightarrow~K.E._{max}=2.4\times10^{-19}~J.$$