An alpha-particle of 5 ; MeV energy strikes a | vedclass

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Question

(C) At the distance of closest approach,the initial kinetic energy of the $\alpha$-particle is entirely converted into electrostatic potential energy.

Vedclass · Accepted Answer

$10^{-12} \; cm$

Vedclass · Answer

(C) At the distance of closest approach,the initial kinetic energy of the $\alpha$-particle is entirely converted into electrostatic potential energy.$K.E. = P.E.$$5 \; MeV = \frac{1}{4\pi\epsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$Given:$K.E. = 5 	imes 10^6 	imes 1.6 	imes 10^{-19} \; J = 8 	imes 10^{-13} \; J$$Z = 92$ (for Uranium)$e = 1.6 	imes 10^{-19} \; C$$\frac{1}{4\pi\epsilon_0} = 9 	imes 10^9 \; N \cdot m^2/C^2$Substituting the values:$8 	imes 10^{-13} = \frac{9 	imes 10^9 	imes 92 	imes 2 	imes (1.6 	imes 10^{-19})^2}{r_0}$Solving for $r_0$:$r_0 = \frac{9 	imes 10^9 	imes 184 	imes 2.56 	imes 10^{-38}}{8 	imes 10^{-13}}$$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 $5 \; MeV$ energy strikes a stationary uranium nucleus at a scattering angle of $180^o$. The distance of closest approach of the $\alpha$-particle to the nucleus will be of the order of:

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