JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2013
- Hard
- MCQ
In an experiment on photoelectric effect, a student plots stopping potential \(V_0\) against reciprocal of the wavelength \(\lambda \) of the incident light for two different metals \(A\) and \(B\). These are shown in the figure Looking at the graphs, you can most appropriately say that
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) Students data is not correct
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{hc}}{\lambda}-\phi=\mathrm{eV}_{0}\)
\(v_{0}=\frac{h c}{e \lambda}-\frac{\phi}{e}\)
For metal \(A\) For metal \(B\)
\(\frac{{\phi A}}{{hc}} = \frac{1}{\lambda }\) \(\frac{{\phi B}}{{hc}} = \frac{1}{\lambda }\)
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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