Question 186 of 200
In an experiment on photoelectric effect, a student plots stopping potential against reciprocal of the wavelength of the incident light for two different metals and . These are shown in the figure.
Looking at the graphs, you can most appropriately say that:
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Answer: (D) 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 (increasing and decreasing) is not specified hence we cannot say that which metal has comparatively greater or lesser work function .
& \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 (increasing and decreasing) is not specified hence we cannot say that which metal has comparatively greater or lesser work function .
%20Work%20function%20of%20metal%20B%20is%20greater%20than%20that%20of%20metal%20A%0A(B)%20For%20light%20of%20certain%20wavelength%20falling%20on%20both%20metal%2C%20maximum%20kinetic%20energy%20of%20electrons%20emitted%20from%20A%20will%20be%20greater%20than%20those%20emitted%20from%20B.%0A(C)%20Work%20function%20of%20metal%20A%20is%20greater%20than%20that%20of%20metal%20B%0A(D)%20Students%20data%20is%20not%20correct){kind=link}