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Physics · 2025 · 2 marks
CBSE 2025 · Region 5 · Set 3 · Q19
Define the term, 'distance of closest approach'. A proton of 3.95 MeV energy approaches a target nucleus Z =79 in head-on position. Calculate its…
Define the term, 'distance of closest approach'. A proton of $\displaystyle 3.95$ MeV energy approaches a target nucleus $\displaystyle \mathrm{Z}=79$ in head-on position. Calculate its distance of closest approach.
Marking-scheme solution
It is the distance from the nucleus at which the $\displaystyle \alpha$ particle stops momentarily and then begins to retrace its path.
It is the distance from the nucleus at which the entire initial kinetic energy of the $\displaystyle \alpha$ particle gets converted into electrostatic potential energy.
$\displaystyle r_{0}=\frac{1}{4 \pi \varepsilon_{0}} \frac{z e^{2}}{K . E}$
$\displaystyle =\frac{9 \times 10^{9} \times 79 \times\left(1.6 \times 10^{-19}\right)^{2}}{3.95 \times 10^{6} \times 1.6 \times 10^{-19}}$
$\displaystyle =28.8 \times 10^{-15} \mathrm{~m}$
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