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

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(B) The distance of closest approach $r_0$ is the point where the initial kinetic energy of the $\alpha$-particle is completely converted into electro

Vedclass · Accepted Answer

$2.88$

Vedclass · Answer

(B) The distance of closest approach $r_0$ is the point where the initial kinetic energy of the $\alpha$-particle is completely converted into electrostatic potential energy.$r_0 = \frac{k q_1 q_2}{K}$Here,$q_1 = 2e$ (charge of $\alpha$-particle),$q_2 = Ze$ (charge of nucleus),and $K = 5 	ext{ MeV} = 5 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J}$.$r_0 = \frac{9 	imes 10^9 	imes (2e) 	imes (Ze)}{K} = \frac{9 	imes 10^9 	imes 2 	imes 50 	imes e^2}{5 	imes 10^6 	imes 1.6 	imes 10^{-19}}$$r_0 = \frac{9 	imes 10^9 	imes 100 	imes (1.6 	imes 10^{-19})^2}{5 	imes 10^6 	imes 1.6 	imes 10^{-19}}$$r_0 = \frac{9 	imes 10^9 	imes 100 	imes 1.6 	imes 10^{-19}}{5 	imes 10^6} = \frac{1440 	imes 10^{-10}}{5 	imes 10^6} = 288 	imes 10^{-16} 	ext{ m} = 2.88 	imes 10^{-14} 	ext{ m}$.Thus,the distance is $2.88 	imes 10^{-14} 	ext{ m}$.

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