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If $K_1$ and $K_2$ are maximum kinetic energies of photoelectrons emitted when lights of wavelengths $\lambda_1$ and $\lambda_2$ respectively incident on a metallic surface. If $\lambda_1=3 \lambda_2$, then
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Verified Answer
The correct answer is:
$K_1 < \left(K_2 / 3\right)$
$K_1=\frac{h c}{\lambda_1}-\phi_0 \ldots$..(i)
$$
\begin{aligned}
& \text { and } \begin{aligned}
K_2=\frac{h c}{\lambda_2}-\phi_0 \text { or } \frac{h c}{\lambda_2}=\left(K_2+\phi_0\right) \\
\begin{aligned}
\therefore \quad K_1-K_2 & =h c\left[\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right] \\
& =h c\left[\frac{1}{3 \lambda_2}-\frac{1}{\lambda_2}\right]=\frac{2 h c}{3 \lambda_2} \\
& =-\frac{2}{3}\left(K_2+\phi_0\right)
\end{aligned} \\
\text { or } K_1=K_2-\frac{2}{3} K_2-\frac{2}{3} \phi_0=\frac{K_2}{3}-\frac{2}{3} \phi_0 \\
\text { or } K_1 < \frac{K_2}{3}
\end{aligned}
\end{aligned}
$$
From (ii)
$$
\begin{aligned}
& \text { and } \begin{aligned}
K_2=\frac{h c}{\lambda_2}-\phi_0 \text { or } \frac{h c}{\lambda_2}=\left(K_2+\phi_0\right) \\
\begin{aligned}
\therefore \quad K_1-K_2 & =h c\left[\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right] \\
& =h c\left[\frac{1}{3 \lambda_2}-\frac{1}{\lambda_2}\right]=\frac{2 h c}{3 \lambda_2} \\
& =-\frac{2}{3}\left(K_2+\phi_0\right)
\end{aligned} \\
\text { or } K_1=K_2-\frac{2}{3} K_2-\frac{2}{3} \phi_0=\frac{K_2}{3}-\frac{2}{3} \phi_0 \\
\text { or } K_1 < \frac{K_2}{3}
\end{aligned}
\end{aligned}
$$
From (ii)
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