When a light of wavelength ' \(\lambda\) ' falls on the emitter of a photocells, maximum speed of emitted photoelectrons is ' \(\mathrm{V}\). If the incident wavelength is changed to \(\frac{2 \lambda}{3}\), maximum speed of emitted photoelectrons will be

\(
(\mathrm{K} . \mathrm{E} .)_1=\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}_0
\)
Multiplying by \(3 / 2\)
\(
\begin{aligned}
& \frac{3}{2}(\mathrm{~K} . \mathrm{E} .)_1=\frac{3}{2} \frac{\mathrm{hc}}{\lambda_1}-\frac{3}{2} \mathrm{~W}_0 ...(1)\\
& (\mathrm{~K} . \mathrm{E} .)_2=\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}_0 \\
& (\mathrm{~K} . \mathrm{E} .)_2=\frac{3}{2} \frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}_0 ...(2)
\end{aligned}
\)
By equation (1) and (2)
\(
\begin{aligned}
& (\mathrm{K} \cdot \mathrm{E} \cdot)_2=\frac{3}{2}(\mathrm{~K} \cdot \mathrm{E} \cdot)_1+\frac{1}{2} \mathrm{~W}_0 \\
& \text { or } \frac{1}{2} \mathrm{mv}_2^2>\frac{3}{2}\left(\frac{1}{2} \mathrm{mv}_1^2\right) \\
& \therefore \mathrm{v}_2^2>\frac{3}{2} \mathrm{v}_1^2 \\
& \therefore \mathrm{v}_2>(1.5)^{1 / 2} \mathrm{v}_1
\end{aligned}
\)