(I). When the frequency of the Incident IIght Is doubled, the kinetic energy of emitted photoelectron becomes more than double.
Let \(E_{1}\) and \(E_{2}\) be the K.E. of photoelectrons for incident light of frequency \(v\) and 2 respectively.
Then,
$$
\begin{aligned}
&h v=E_{1}+\phi_{0} \text { and } \\
&2 h v=E_{2}-\phi_{0} \\
&\therefore 2=\frac{E_{2}+\phi_{0}}{E_{2}+\phi_{0}}
\end{aligned}
$$
$$
\begin{aligned}
&O r, 2 E_{1}+2 \phi_{0}=E_{2}-\phi_{0} \\
&O r, E_{2}+2 E_{1}+\phi_{0} \\
&O r, E_{2}=2 E_{1}
\end{aligned}
$$
(II). Photoelectric current remains unchanged as It depends upon the Intensity of Incldent light.
(III). The stopping potentlal becomes more than double.
$$
\begin{aligned}
&\mathrm{eV}_{0}=(\mathrm{K} . \mathrm{E} .)_{\max }=\mathrm{hv}-\phi_{0} \\
&o r_{1} \mathrm{hv}=\mathrm{e}_{0}+\phi_{0} \text { and } \\
&\mathrm{eV}_{0}^{\prime}=2 \mathrm{hv}-\phi_{0} \\
&=2\left(\mathrm{eV}_{0}-\phi_{0}\right)-\phi_{0} \\
&=2 \mathrm{eV}_{0}-\phi_{0} \\
&\therefore \mathrm{V}_{0}^{3}=2_{0}
\end{aligned}
$$
