When a certain metal was irradiated with light of frequency $4.0 \times 10^{16} \mathrm{~s}^{-1}$, the…
When a certain metal was irradiated with light of frequency $4.0 \times 10^{16} \mathrm{~s}^{-1}$, the photoelectrons emitted had four times kinetic energy as the kinetic energy of photoelectrons emitted when the same metal was irradiated with light of frequency $2.0 \times 10^{16} \mathrm{~s}^{-1}$. The threshold frequency $\left(v_0\right)$ of the metal in $\mathrm{s}^{-1}$ is
$2 \times 10^{16}$
$4 \times 10^{16}$
$2.5 \times 10^{16}$
$1.33 \times 10^{16}$
Solution
Given,
$
\begin{aligned}
v_1 & =4 \times 10^{16} \mathrm{sec}^{-1} \\
v_2 & =2 \times 10^{16} \mathrm{sec}^{-1} \\
\mathrm{KE}_1 & =4 \mathrm{KE}_2
\end{aligned}
$
From equation of photoelectric effect,
On putting the above calculated value in eq (ii), we get
$\begin{gathered}\frac{2 h v_2}{3}=h v_0 \\ \frac{2 \times h \times 2 \times 10^{16}}{3}=h v_0 \\ v_0=1.33 \times 10^{16}\end{gathered}$
