CBSE 2025 · Region 2 · Set 3 · Q20 · 2 marks
The threshold wavelength of a metal is $\displaystyle 450$ nm . Calculate (i) the work function of the metal in eV and (ii) the maximum energy of the ejected photoelectrons in eV by incident radiation of $\displaystyle 250$ nm .
Marking-scheme solution
(i)
$\displaystyle \phi=\frac{h c}{\lambda_{0}}$
$\displaystyle \phi=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{450 \times 10^{-9} \times 1.6 \times 10^{-19}}$
$\displaystyle \phi=2.76 \mathrm{~eV}$
(ii)
$\displaystyle K_{\max }=h \nu-\phi$
$\displaystyle h \nu=\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{250 \times 10^{-9} \times 1.6 \times 10^{-19}}=4.97 \mathrm{~eV}$
$\displaystyle K_{\max }=4.97-2.76$
$\displaystyle K_{\max }=2.21 \mathrm{~eV}$
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CBSE Class 12 Physics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.