Question

The maximum kinetic energy of a photoelectron is $3eV$ . Its stopping potential is

• A. $3V$
• B. $1V$
• C. $6V$
• D. $2V$

Answer 1

Answer: (A)

Here,
$K_{max}=3\text{eV}$
Maximum kinetic energy is given by:
$$\begin{gathered} K_{max}=e{V}_0 \\ 3eV=e\times V_0 \\ 3V=V_0 \end{gathered}$$
$\therefore$ Stopping potential $V_0=3V$