The threshold frequency of a photoelectric metal is $v_0$. If light of frequency $4 v_0$ is incident on this…
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:
$h v_0$
$2 h v_0$
$3 h v_0$
$4 h \mathrm{v}_0$
Solution
By using Einstein's photoelectric equation, we have,
$\begin{aligned}
& \mathrm{KE}=h v-\phi \\
& \mathrm{KE}=h v-h v_0 \\
& =\mathrm{h}\left(4 v_0\right)-h v_0 \\
& =3 h v_0 \\
& \text { where, } \quad h=\text { Planck's constant, } \\
& v \& v_0=\text { frequencies, and } \\
& \phi=\text { work function. }
\end{aligned}$