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The threshold frequency of a photoelectric metal is $v_0$. If light of frequency $4 v_0$ is incident on this metal, then the maximum kinetic energy of emitted electrons will be:
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Verified Answer
The correct answer is:
$3 h v_0$
By using Einstein's photoelectric equation, we have,
$\begin{aligned}
& \mathrm{KE}=h v-\phi \\
& \mathrm{KE}=h v-h v_0 \\
& =\mathrm{h}\left(4 v_0\right)-h v_0 \\
& =3 h v_0 \\
& \text { where, } \quad h=\text { Planck's constant, } \\
& v \& v_0=\text { frequencies, and } \\
& \phi=\text { work function. } \\
&
\end{aligned}$
$\begin{aligned}
& \mathrm{KE}=h v-\phi \\
& \mathrm{KE}=h v-h v_0 \\
& =\mathrm{h}\left(4 v_0\right)-h v_0 \\
& =3 h v_0 \\
& \text { where, } \quad h=\text { Planck's constant, } \\
& v \& v_0=\text { frequencies, and } \\
& \phi=\text { work function. } \\
&
\end{aligned}$
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