Question

Two identical photocathodes receive light of frequencies \(f_{1}\) and \(f_{2}\). If the velocities of the photoelectrons (of mass \(m\) ) coming out are respectively \(v_{1}\) and \(v_{2}\), then
(a) \(v_{1}{ }^{2}-v_{2}^{2}=\frac{2 h}{m}\left(f_{1}-f_{2}\right)\)
(b) \(v_{1}+v_{2}=\left[\frac{2 h}{m}\left(f_{1}+f_{2}\right)\right]^{\frac{1}{2}}\)
(c) \(v_{1}^{2}+v_{2}^{2}=\frac{2 h}{m}\left(f_{1}+f_{2}\right)\)
(d) \(v_{1}-v_{2}=\left[\frac{2 h}{m}\left(f_{1}-f_{2}\right)\right]^{\frac{1}{2}}\).

Answer 1

The correct answer is:
(a): For photoelectric effect, according to Einstein's equation,
Kinetic energy of emitted electron \(=\) \(h f-(\) work function \(\phi)\)
$$
\begin{aligned}
\therefore \quad & \frac{1}{2} m v_{1}^{2}=h f_{1}-\phi \\
& \frac{1}{2} m v_{2}^{2}=h f_{2}-\phi \\
\therefore \quad & \frac{1}{2} m\left(v_{1}^{2}-v_{2}^{2}\right)=h\left(f_{1}-f_{2}\right) \\
\therefore \quad & v_{1}^{2}-v_{2}^{2}=\frac{2 h}{m}\left(f_{1}-f_{2}\right) .
\end{aligned}
$$