CBSE 2023 · Region 4 · Set 1 · Q23 · 2 marks
The figure shows $\displaystyle \mathrm{v}_{\mathrm{m}}^{2}$ versus $\displaystyle \frac{1}{\lambda}$ graph for photoelectrons emitted from a surface where $\displaystyle \mathrm{v}_{\mathrm{m}}$ is the maximum speed of electrons and $\displaystyle \lambda$ is the wavelength of incident radiation. Using this graph and Einstein's photoelectric equation, obtain the expression for Planck's constant and work function of the surface.

Marking-scheme solution
\[\begin{aligned}
& \frac{1}{2} \mathrm{m} \mathrm{v}_{\mathrm{m}}^{2}=\frac{\mathrm{h} c}{\lambda}-\phi_{0} \\
& \mathrm{v}_{\mathrm{m}}^{2}=\left(\frac{2 \mathrm{h} c}{\mathrm{m}}\right) \frac{1}{\lambda}-\frac{2}{\mathrm{m}} \phi_{0}
\end{aligned}
\]
According to this equation a plot of \(\displaystyle \mathrm{v}_{\mathrm{m}}^{2}\) versus \(\displaystyle (1 / \lambda)\) is a straight line.
Slope of the graph \(\displaystyle =\frac{2 \mathrm{h} c}{\mathrm{m}}\)
\[\text { Intercept }=\frac{2}{\mathrm{m}} \phi_{0}
\]
Slope and intercept can be found from the graph
\[\begin{aligned}
& \mathrm{h}=\frac{\mathrm{m}}{2 c} \times \text { slope } \\
& \phi_{0}=\frac{\mathrm{m}}{2} \times \text { intercept }
\end{aligned}
\]
Dual Nature of Radiation and MatterEinstein’s Photoelectric Equation: Energy Quantum of RadiationApplyvery_short_answermedium
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CBSE Class 12 Physics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.