If an alpha particle with energy 7.7 text{ Me | vedclass

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(A) The distance of closest approach $r_0$ is the distance where the kinetic energy of the alpha particle is completely converted into electrostatic p

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$2.95 	imes 10^{-14}$

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(A) The distance of closest approach $r_0$ is the distance where the kinetic energy of the alpha particle is completely converted into electrostatic potential energy.Using the principle of conservation of energy:$K_i + U_i = K_f + U_f$At the distance of closest approach,final kinetic energy $K_f = 0$.$K_i = \frac{1}{4 \pi \epsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$Given:$K_i = 7.7 	ext{ MeV} = 7.7 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J}$$Z = 79$ (for gold)$e = 1.6 	imes 10^{-19} 	ext{ C}$$\frac{1}{4 \pi \epsilon_0} = 9 	imes 10^9 	ext{ N m}^2/	ext{C}^2$Substituting the values:$7.7 	imes 10^6 	imes 1.6 	imes 10^{-19} = \frac{9 	imes 10^9 	imes (79 	imes 1.6 	imes 10^{-19}) 	imes (2 	imes 1.6 	imes 10^{-19})}{r_0}$$r_0 = \frac{9 	imes 10^9 	imes 79 	imes 2 	imes 1.6 	imes 10^{-19}}{7.7 	imes 10^6}$$r_0 \approx 2.95 	imes 10^{-14} 	ext{ m}$

If an alpha particle with energy $7.7 \text{ MeV}$ is bombarded on a thin gold foil,the closest distance from the nucleus it can reach is . . . . . . $\text{m}$. (Atomic number of gold $= 79$ and $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9$ in $SI$ units)

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