CBSE 2024 · Region 5 · Set 3 · Q23 · 3 marks
The threshold frequency for a metal is $\displaystyle 3.0 \times 10^{14} \mathrm{~Hz}$. A beam of frequency $\displaystyle 9.0 \times 10^{14} \mathrm{~Hz}$ is incident on the metal. Calculate (i) the work function (in eV ) of the metal and (ii) the maximum speed of photoelectrons.
Marking-scheme solution
(i)
$\displaystyle \phi_{0}=h \nu_{0}=\frac{6.63 \times 10^{-34} \times 3.0 \times 10^{14}}{1.6 \times 10^{-19}}$
$\displaystyle =1.24 \mathrm{~eV}$
(ii)
$\displaystyle K_{\max }=h \nu-h \nu_{0}$
$\displaystyle \frac{1}{2} m V_{\max }^{2}=h\left(\nu-\nu_{0}\right)$
$\displaystyle V_{\max }=\left[\frac{2 h\left(\nu-\nu_{0}\right)}{m}\right]^{\frac{1}{2}}$
$\displaystyle V_{\max }=\left[\frac{2 \times 6.63 \times 10^{-34}(9-3) \times 10^{14}}{9.1 \times 10^{-31}}\right]^{\frac{1}{2}}$
$\displaystyle =9.35 \times 10^{5} \mathrm{~m} / \mathrm{s}$
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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.