A 7.9 text{ MeV} alpha-particle scatters from | vedclass

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(A) At the distance of closest approach $(r)$, the entire kinetic energy of the $\alpha$-particle is converted into electrostatic potential energy.By

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$5.76 	imes 10^{-14}$

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(A) At the distance of closest approach $(r)$, the entire kinetic energy of the $\alpha$-particle is converted into electrostatic potential energy.By the principle of conservation of mechanical energy:$K.E._i + P.E._i = K.E._f + P.E._f$Given $K.E._i = 7.9 	ext{ MeV} = 7.9 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J}$, $P.E._i = 0$, $K.E._f = 0$, and $P.E._f = \frac{1}{4 \pi \epsilon_0} \frac{(2e)(Ze)}{r}$.$7.9 	imes 10^6 	imes 1.6 	imes 10^{-19} = \frac{9 	imes 10^9 	imes 2 	imes 79 	imes (1.6 	imes 10^{-19})^2}{r}$Solving for $r$:$r = \frac{9 	imes 10^9 	imes 2 	imes 79 	imes 1.6 	imes 10^{-19}}{7.9 	imes 10^6} = 2.88 	imes 10^{-14} 	ext{ m}$The diameter of the nucleus is $D = 2r = 2 	imes 2.88 	imes 10^{-14} = 5.76 	imes 10^{-14} 	ext{ m}$.

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