Question

Light of frequency $4v_0$ is incident on the metal of the threshold frequency $v_0$. The maximum kinetic energy of the emitted photoelectrons is

• A. $3h{v}_0$
• B. $2h{v}_0$
• C. $\frac{3}{2}h{v}_0$
• D. $\frac{1}{2}h{v}_0$

Answer 1

The correct option is A

$3h{v}_0$

$E=h{v}_0+K_{max}\Rightarrow h\left(4v_0\right)=h{v}_0+K_{max}\Rightarrow K_{max}=3h{v}_0$