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In an experiment on photoelectric effect, a student plots stopping potential $\mathrm{V}_0$ against reciprocal of the wavelength $\lambda$ of the incident light for two different metals $\mathrm{A}$ and $\mathrm{B}$. These are shown in the figure.
Looking at the graphs, you can most appropriately say that:
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Looking at the graphs, you can most appropriately say that:
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Verified Answer
The correct answer is:
Students data is not correct
Students data is not correct
$\begin{aligned}
& \frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{eV}_0 \\
& \mathrm{v}_0=\frac{\mathrm{hc}}{\mathrm{e} \lambda}-\frac{\phi}{\mathrm{e}}
\end{aligned}$
As the value of $\frac{1}{\lambda}$ (increasing and decreasing) is not specified hence we cannot say that which metal has comparatively greater or lesser work function $(\phi)$.
& \frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{eV}_0 \\
& \mathrm{v}_0=\frac{\mathrm{hc}}{\mathrm{e} \lambda}-\frac{\phi}{\mathrm{e}}
\end{aligned}$
As the value of $\frac{1}{\lambda}$ (increasing and decreasing) is not specified hence we cannot say that which metal has comparatively greater or lesser work function $(\phi)$.
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