A particle of mass $M$ , charge $q > 0$ and initial kinetic energy $K$ is projected from infinity towards a heavy nucleus of charge $Q$ assumed to have a fixed position.
(a) If the aim is perfect, how close to the center of the nucleus is the particle when it comes instantaneously to rest?
(b) With a particular imperfect aim, the particle's closest approach to nucleus is twice the distance determined in (a) Determine speed of particle at the closest distance of approach.

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Solution

a.

when the particle comes to rest instantaneously near the nucleolus its total energy is in the form of potential energy

$= \frac{1}{4 π ϵ} \frac{Q q}{r}$

total kinetic energy is given by,

$K E = \frac{1}{4 π ϵ} \frac{Q q}{r}$

$\Rightarrow r = \frac{1}{4 π ϵ} \frac{Q q}{K E}$

b.

let the distance of closest approach be,

$d = 2 r$

$d = \frac{1}{4 π ϵ} \frac{2 Q q}{K E}$

potential energy of the particle when at d

$= \frac{1}{4 π ϵ} \frac{Q q}{d}$

$= \frac{1}{4 π ϵ} Q q \times \frac{4 π ϵ K E}{2 Q q} = \frac{K E}{2}$

The kinetic energy is, $K E - \frac{K E}{2}$

If v is the velocity at the closest distance of approach,

$\frac{1}{2} m v^{2} = \frac{K E}{2}$

$v = \sqrt{\frac{K E}{m}}$

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