CBSE 2024 · Region 5 · Set 2 · Q22 · 3 marks
A photosensitive surface of work function $\displaystyle 2.1$ eV is irradiated by radiation of wavelength $\displaystyle 150$ nm . Calculate (i) the threshold wavelength, (ii) energy (in eV ) of an incident photon, and (iii) maximum kinetic energy of emitted photoelectron.
Marking-scheme solution
$\displaystyle \phi_{0}=\frac{h c}{\lambda_{0}}$
(i)
$\displaystyle \lambda_{0}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{2.1 \times 1.6 \times 10^{-19}}$
$\displaystyle =5.92 \times 10^{-7} \mathrm{~m}$
(ii)
Energy of incident photon $\displaystyle =\frac{h c}{\lambda}$
$\displaystyle =\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{150 \times 10^{-9} \times 1.6 \times 10^{-19}}$
$\displaystyle =8.29 \mathrm{~eV}$
(iii)
Using Einstein's equation:
$\displaystyle \frac{h c}{\lambda}=\phi_{0}+K_{\max }$
$\displaystyle K_{\max }=(8.29-2.1) \mathrm{eV}$
$\displaystyle =6.2 \mathrm{~eV}$
Dual Nature of Radiation and MatterEinstein’s Photoelectric Equation: Energy Quantum of RadiationApplyshort_answermedium
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.