An elementary particle of mass m and charge + | vedclass

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Question

(A) According to the law of conservation of energy,the initial kinetic energy of the incident particle is converted into electrostatic potential energ

Vedclass · Accepted Answer

$\frac{Ze^2}{2\pi \varepsilon_0 mv^2}$

Vedclass · Answer

(A) According to the law of conservation of energy,the initial kinetic energy of the incident particle is converted into electrostatic potential energy at the point of closest approach $(r)$.Initial energy = Final energy$\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(Ze)(e)}{r}$Rearranging for $r$:$r = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Ze^2}{\frac{1}{2}mv^2}$$r = \frac{2Ze^2}{4\pi\varepsilon_0 mv^2} = \frac{Ze^2}{2\pi\varepsilon_0 mv^2}$

An elementary particle of mass $m$ and charge $+e$ is projected with velocity $v$ at a much more massive particle of charge $Ze$,where $Z > 0$. What is the closest possible approach of the incident particle?

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