An alpha-particle of energy 4 text{ MeV} is s | vedclass

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Question

(C) At the distance of closest approach $(r_0)$, the entire initial kinetic energy of the $\alpha$-particle is converted into electrostatic potential

Vedclass · Accepted Answer

$10^{-12} 	ext{ cm}$

Vedclass · Answer

(C) At the distance of closest approach $(r_0)$, the entire initial kinetic energy of the $\alpha$-particle is converted into electrostatic potential energy.Given: Kinetic energy $K = 4 	ext{ MeV} = 4 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J} = 6.4 	imes 10^{-13} 	ext{ J}$.Atomic number of uranium $(Z)$ = $92$.Atomic number of $\alpha$-particle $(z)$ = $2$.The formula for the distance of closest approach is $r_0 = \frac{1}{4\pi\epsilon_0} \cdot \frac{(ze)(Ze)}{K}$.Substituting the values:$r_0 = \frac{9 	imes 10^9 	imes (2 	imes 1.6 	imes 10^{-19}) 	imes (92 	imes 1.6 	imes 10^{-19})}{6.4 	imes 10^{-13}}$$r_0 = \frac{9 	imes 10^9 	imes 2 	imes 92 	imes (1.6 	imes 10^{-19})^2}{6.4 	imes 10^{-13}}$$r_0 = \frac{9 	imes 10^9 	imes 184 	imes 2.56 	imes 10^{-38}}{6.4 	imes 10^{-13}}$$r_0 = \frac{4239.36 	imes 10^{-29}}{6.4 	imes 10^{-13}} \approx 662.4 	imes 10^{-16} 	ext{ m} = 6.624 	imes 10^{-14} 	ext{ m}$.Converting to centimeters: $r_0 = 6.624 	imes 10^{-14} 	imes 10^2 	ext{ cm} = 6.624 	imes 10^{-12} 	ext{ cm}$.This is of the order of $10^{-12} 	ext{ cm}$.

An $\alpha$-particle of energy $4 \text{ MeV}$ is scattered through $180^\circ$ by a fixed uranium nucleus. The distance of the closest approach is of the order of

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