An alpha-particle of energy 5 , MeV is scatte | vedclass

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(C) The distance of closest approach $r_0$ is the distance where the initial kinetic energy of the $\alpha$-particle is entirely converted into electr

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$10^{-12} \, cm$

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(C) The distance of closest approach $r_0$ is the distance where the initial kinetic energy of the $\alpha$-particle is entirely converted into electrostatic potential energy.$K.E. = P.E. = \frac{1}{4\pi\epsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$Given: $K.E. = 5 \, MeV = 5 	imes 10^6 	imes 1.6 	imes 10^{-19} \, J = 8 	imes 10^{-13} \, J$.For Uranium $(Z = 92)$,$k = 9 	imes 10^9 \, N \cdot m^2/C^2$.$r_0 = \frac{k \cdot (Ze) \cdot (2e)}{K.E.} = \frac{9 	imes 10^9 	imes 92 	imes 2 	imes (1.6 	imes 10^{-19})^2}{8 	imes 10^{-13}}$$r_0 \approx 5.3 	imes 10^{-14} \, m = 5.3 	imes 10^{-12} \, cm$.Thus,the order of magnitude is $10^{-12} \, cm$.

An $\alpha$-particle of energy $5 \, MeV$ is scattered by a fixed uranium nucleus at $180^o$. What is the distance of closest approach between the particle and the uranium nucleus?

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