An alpha-particle of energy 5 text{ MeV} is s | vedclass

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Question

(B) The distance of closest approach $(d)$ is the distance at which the kinetic energy of the $\alpha$-particle is completely converted into electrost

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

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(B) The distance of closest approach $(d)$ is the distance at which the kinetic energy of the $\alpha$-particle is completely converted into electrostatic potential energy.$d = \frac{1}{4 \pi \varepsilon_{0}} \frac{(Z_1 e)(Z_2 e)}{K}$Here,$Z_1 = 2$ (for $\alpha$-particle),$Z_2 = 79$ (for gold nucleus),and $K = 5 	ext{ MeV} = 5 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J} = 8 	imes 10^{-13} 	ext{ J}$.Substituting the values:$d = (9 	imes 10^9) 	imes \frac{(2 	imes 1.6 	imes 10^{-19}) 	imes (79 	imes 1.6 	imes 10^{-19})}{8 	imes 10^{-13}}$$d = \frac{9 	imes 10^9 	imes 2 	imes 79 	imes 2.56 	imes 10^{-38}}{8 	imes 10^{-13}}$$d \approx 4.55 	imes 10^{-14} 	ext{ m} = 4.55 	imes 10^{-12} 	ext{ cm}$.Thus,the order of magnitude is $10^{-12} 	ext{ cm}$.

An $\alpha$-particle of energy $5 \text{ MeV}$ is scattered through $180^{\circ}$ by a gold nucleus. The distance of closest approach is of the order of:

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