In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 imes 10^{-14} \mathrm{~m}$. If target nucleus has atomic number 80 , then maximum velocity of $\alpha$ - particle is _______ $ imes 10^5 \mathrm{~m} / \mathrm{s}$ approximately.…

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In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 	imes 10^{-14} \mathrm{~m}$. If target nucleus has atomic number 80 , then maximum velocity of $\alpha$ - particle is _______ $	imes 10^5 \mathrm{~m} / \mathrm{s}$ approximately. $\left(\frac{1}{4 \pi \epsilon_0}=9 	imes 10^9ight.$ SI unit, mass of $\alpha$ particle $\left.=6.72 	imes 10^{-27} \mathrm{~kg}ight)$

Aicharya · Accepted Answer

$\begin{aligned} v & =\sqrt{\frac{4 \mathrm{KZe}^2}{\mathrm{mr}_{	ext {min }}}} \ & =\sqrt{\frac{4 	imes 9 	imes 10^9 	imes 80}{6.72 	imes 10^{-27} 	imes 4.5 	imes 10^{-14}}} 	imes 1.6 	imes 10^{-19} \ & =9.759 	imes 10^{25} 	imes 1.6 	imes 10^{-19} \ & =156 	imes 10^5 \mathrm{~m} / \mathrm{s}\end{aligned}$

In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5…

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