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In a photoelectric experiment, with light of wavelength $\lambda,$ the fastest electron has speed $v$. If the exciting wavelength is changed to $3 \lambda / 4$, the speed of the fastest emitted electron will become
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greater than $v \sqrt{\frac{4}{3}}$
$\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{hc}}{\lambda}-\phi$
$$
\begin{array}{l}
\frac{1}{2} \mathrm{~m}^{\prime}=\frac{\mathrm{hc}}{(3 \lambda / 4)}-\phi=\frac{4 \mathrm{hc}}{3 \lambda}-\phi \\
\text { Clearly, } \mathrm{v}^{\prime}>\sqrt{\frac{4}{3} \mathrm{v}}
\end{array}
$$
$$
\begin{array}{l}
\frac{1}{2} \mathrm{~m}^{\prime}=\frac{\mathrm{hc}}{(3 \lambda / 4)}-\phi=\frac{4 \mathrm{hc}}{3 \lambda}-\phi \\
\text { Clearly, } \mathrm{v}^{\prime}>\sqrt{\frac{4}{3} \mathrm{v}}
\end{array}
$$
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