alpha-particles of energy 400 , KeV are bomba | vedclass

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(D) The minimum distance of approach $(r_0)$ occurs when the kinetic energy of the $\alpha$-particle is completely converted into electrostatic potent

Vedclass · Accepted Answer

$0.59 \, pm$

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(D) The minimum distance of approach $(r_0)$ occurs when the kinetic energy of the $\alpha$-particle is completely converted into electrostatic potential energy at the point of closest approach.Given:Kinetic Energy $(K)$ = $400 \, KeV = 400 	imes 10^3 	imes 1.6 	imes 10^{-19} \, J = 6.4 	imes 10^{-14} \, J$.Atomic number of Lead $(Z)$ = $82$.Charge of $\alpha$-particle $(q_1)$ = $2e = 2 	imes 1.6 	imes 10^{-19} \, C$.Charge of Lead nucleus $(q_2)$ = $Ze = 82 	imes 1.6 	imes 10^{-19} \, C$.Using the conservation of energy:$K = \frac{1}{4\pi\epsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$$6.4 	imes 10^{-14} = (9 	imes 10^9) \cdot \frac{82 	imes 2 	imes (1.6 	imes 10^{-19})^2}{r_0}$$r_0 = \frac{9 	imes 10^9 	imes 164 	imes 2.56 	imes 10^{-38}}{6.4 	imes 10^{-14}}$$r_0 = \frac{3774.912 	imes 10^{-29}}{6.4 	imes 10^{-14}} \approx 5.9 	imes 10^{-13} \, m$$r_0 = 0.59 	imes 10^{-12} \, m = 0.59 \, pm$.

$\alpha$-particles of energy $400 \, KeV$ are bombarded on a nucleus of $_{82}Pb$. In the scattering of $\alpha$-particles, what will be the minimum distance of approach from the nucleus?

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