Electrostatics

JEE Main level, single correct, medium

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Question 231 of 311

An α\alpha-particle of energy 5MeV5 \mathrm{MeV} is scattered through 180∘180^{\circ} by a fixed uranium nucleus. The distance of the closest approach is of the order of

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Answer: (C) 10−12 cm10^{-12} \mathrm{~cm}

At closest approach, all the kinetic energy of the α\alpha-particle will converted into the potential energy of the system, K.E. = P.E.
5MeV=14πε0q1q2r5×106×e=9×109Z1×Z2e2rr=9×109×92×2×1.6×10−195×106∴r=5.3×10−14 m=5.3×10−12 cm\begin{aligned} & 5 \mathrm{MeV}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}_1 \mathrm{q}_2}{\mathrm{r}} \\ & 5 \times 10^6 \times \mathrm{e}=9 \times 10^9 \frac{\mathrm{Z}_1 \times \mathrm{Z}_2 \mathrm{e}^2}{\mathrm{r}} \\ & \mathrm{r}=\frac{9 \times 10^9 \times 92 \times 2 \times 1.6 \times 10^{-19}}{5 \times 10^6} \\ & \therefore \mathrm{r}=5.3 \times 10^{-14} \mathrm{~m}=5.3 \times 10^{-12} \mathrm{~cm} \end{aligned}

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