CBSE 2024 · Region 5 · Set 2 · Q21 · 2 marks
A proton of energy $\displaystyle 1.6$ MeV approaches a gold nucleus ( $\displaystyle \mathrm{Z}=79$ ). Find the distance of its closest approach.
Marking-scheme solution
$\displaystyle d_{0}=\frac{k Z e^{2}}{K_{p}}$
$\displaystyle =\frac{9 \times 10^{9} \times 79 \times\left(1.6 \times 10^{-19}\right)^{2}}{1.6 \times 1.6 \times 10^{-19} \times 10^{6}}$
$\displaystyle =711 \times 10^{-16} \mathrm{~m}$
$\displaystyle =7.11 \times 10^{-14} \mathrm{~m}$
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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.