(a) Given, $v=6 \times 10^{14} \mathrm{~Hz}$,
$$
\begin{aligned}
\phi_0 &=2.14 \mathrm{eV} \\
&=2.14 \times 1.6 \times 10^{-19} \mathrm{~J}
\end{aligned}
$$
By Einstein's photoelectric equation,
$$
\begin{aligned}
&\mathrm{K}_{\max }=\frac{1}{2} \mathrm{mv}^2=\mathrm{hv}-\phi_0 \\
&=6.6 \times 10^{-34} \times 6 \times 10^{14}-2.14 \times 1.6 \times 10^{-19} \\
&=3.97 \times 10^{-19}-3.42 \times 10^{-19} \\
&\cong 0.550 \times 10^{-19} \mathrm{~J} . \\
&=\frac{0.55 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}=0.343 \mathrm{eV} .
\end{aligned}
$$
(b) Stopping potential $\mathrm{V}_0=\frac{\mathrm{K}_{\max }}{\mathrm{e}} \mathrm{eV}$ $=\frac{0.343 \mathrm{eV}}{\mathrm{e}}=0.343 \mathrm{~V}$
(c) Maximum speed of the photoelectrons,
$$
\begin{aligned}
\mathrm{V}_{\max } &=\sqrt{\frac{\left(\mathrm{h} v-\phi_0\right) \times 2}{\mathrm{~m}}}(\text { from (a)) }\\
&=\sqrt{\frac{2 \times 0.548 \times 10^{-19}}{9.1 \times 10^{-31}}}\left(\because \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}\right) \\
&=\sqrt{0.12 \times 10^{12}}=3.47 \times 10^5 \mathrm{~m} / \mathrm{s} .
\end{aligned}
$$