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Q.
An $\alpha $ -particle of energy $5 \, MeV$ is scattered through $180^\circ $ by a stationary uranium nucleus. The distance of closest approach is of the order
NTA AbhyasNTA Abhyas 2022
Solution:
According to work energy theorem
$KE=PE$
$KE=\frac{1}{4 \pi \epsilon _{0}}\frac{q_{1} q_{2}}{r}$
$r=\frac{1}{4 \pi \epsilon _{0}}\frac{92 e \cdot 2 e}{E}$
$r=\frac{9 \times 1 0^{9} \times 92 \times 2 \left(1.6 \times 1 0^{- 19}\right)^{2}}{5 \times 1 0^{6} \times \left(1.6 \times 1 0^{- 19}\right)}$
$r=5.3\times 10^{- 14}m \approx10^{- 12} \, cm.$