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Light of frequency $4 \times 10^{14} \mathrm{~Hz}$ is incident on a metal surface of work function $2.14 \mathrm{eV}$, resulting in photoemission of electrons. The maximum kinetic energy of the emitted electrons is $\left[h=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right.$ ]
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The correct answer is:
$0 \mathrm{eV}$
Energy of a light photon, $E=h f$
$$
\begin{aligned}
& =6.63 \times 10^{-34} \times 4 \times 10^{24} \\
& =6.63 \times 4 \times 10^{-20} \mathrm{~J} \\
& =\frac{6.63 \times 4 \times 10^{-20}}{1.6 \times 10^{-29}} \mathrm{eV}=1.65 \mathrm{eV}
\end{aligned}
$$
As, energy of incident photon is less than the work function of metal, no photo emission will takes place.
$$
\begin{aligned}
& =6.63 \times 10^{-34} \times 4 \times 10^{24} \\
& =6.63 \times 4 \times 10^{-20} \mathrm{~J} \\
& =\frac{6.63 \times 4 \times 10^{-20}}{1.6 \times 10^{-29}} \mathrm{eV}=1.65 \mathrm{eV}
\end{aligned}
$$
As, energy of incident photon is less than the work function of metal, no photo emission will takes place.
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