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

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(D) At the distance of closest approach,the entire kinetic energy of the alpha particle is converted into electrostatic potential energy.$K.E. = U$$K

Vedclass · Accepted Answer

$28.8 	imes 10^{-16} \frac{Z}{K}$

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(D) At the distance of closest approach,the entire kinetic energy of the alpha particle is converted into electrostatic potential energy.$K.E. = U$$K 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J} = \frac{1}{4 \pi \epsilon_0} \frac{(2e)(Ze)}{r}$Substituting the value of $\frac{1}{4 \pi \epsilon_0} = 9 	imes 10^9 	ext{ N m}^2/	ext{C}^2$ and $e = 1.6 	imes 10^{-19} 	ext{ C}$:$r = \frac{2 	imes 9 	imes 10^9 	imes Z 	imes (1.6 	imes 10^{-19})^2}{K 	imes 10^6 	imes 1.6 	imes 10^{-19}}$$r = \frac{18 	imes 10^9 	imes 1.6 	imes 10^{-19} 	imes Z}{K 	imes 10^6}$$r = \frac{28.8 	imes 10^{-10}}{K 	imes 10^6} Z$$r = 28.8 	imes 10^{-16} \frac{Z}{K} 	ext{ m}$

An alpha particle of energy $K \text{ MeV}$ is moving towards a nucleus of atomic number $Z$. The distance of closest approach of the alpha particle to the nucleus in metres is

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