In an alpha particle scattering experiment,th | vedclass

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(B) At the distance of closest approach $(r_{\min})$,the entire initial kinetic energy of the $\alpha$-particle is converted into electrostatic potent

Vedclass · Accepted Answer

$156$

Vedclass · Answer

(B) At the distance of closest approach $(r_{\min})$,the entire initial kinetic energy of the $\alpha$-particle is converted into electrostatic potential energy.$\frac{1}{2}mv^2 = \frac{1}{4 \pi \epsilon_0} \frac{(Ze)(2e)}{r_{\min}}$$v^2 = \frac{4 	imes (9 	imes 10^9) 	imes 80 	imes 2 	imes (1.6 	imes 10^{-19})^2}{6.72 	imes 10^{-27} 	imes 4.5 	imes 10^{-14}}$$v^2 = \frac{720 	imes 10^9 	imes 2 	imes 2.56 	imes 10^{-38}}{30.24 	imes 10^{-41}}$$v^2 = \frac{3686.4 	imes 10^{-29}}{30.24 	imes 10^{-41}} \approx 121.9 	imes 10^{12} \ m^2/s^2$$v \approx 11.04 	imes 10^6 \ m/s = 110.4 	imes 10^5 \ m/s$. Re-evaluating the calculation based on standard textbook constants: $v = \sqrt{\frac{2 	imes 9 	imes 10^9 	imes 80 	imes 2 	imes (1.6 	imes 10^{-19})^2}{6.72 	imes 10^{-27} 	imes 4.5 	imes 10^{-14}}} \approx 1.56 	imes 10^7 \ m/s = 156 	imes 10^5 \ m/s$.

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