CBSE 2026 · Region 1 · Set 1 · Q17 · 2 marks
In a photoelectric experiment, the emitter plate is irradiated with radiation of $\displaystyle 200$ nm . The photocurrent becomes zero when the collector plate potential is - $\displaystyle 0.80$ V. Calculate the work function (in eV) of the emitter.
Marking-scheme solution
$\displaystyle \mathrm{h} \nu=\phi_{0}+\mathrm{KE}_{\text {max }}$
$\displaystyle \frac{\mathrm{hc}}{\lambda}=\phi_{0}+\mathrm{eV}_{0}$
$\displaystyle \phi_{0}=\frac{\mathrm{hc}}{\lambda}-\mathrm{eV}_{0}$
$\displaystyle \phi_{0}=\left[\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 200 \times 10^{-9}}-0.80\right] \mathrm{eV}$
$\displaystyle \phi_{0}=5.4 \mathrm{eV}$
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CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.