An - particle collides with nucleus after passing through the pot… - Vidyadip

Question

An $\alpha -$ particle collides with nucleus after passing through the potential $V$ volt. Prove that distance of closest approach of particle of atomic number $Z$ to the nucleus will be $14.4( Z / V ) \ \mathring A $.
$($Given that : $1 / 4 \pi \varepsilon_0=9.0 \times 10^9$ Newton $-$ meter $^2 /$ Coulomb $^2$ and $e=1.6 \times 10^{-19}$ Coulomb$)$

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Answer

Energy of the $\alpha-$ particle to reach the closest to the nucleus :
Kinetic energy of $\alpha-$ particle $=$ Electrostatic potential energy of $\alpha-$ particle and nucleus system.
$2 e \times V=\frac{1}{4 \pi \in_0}\left(\frac{2 e \times Z e }{r_0}\right)$
$=9 \times 10^9\left(\frac{2 e \times Z e }{r_0}\right)$
$\Rightarrow 2 eV=\frac{2 \times 9 \times 10^9 \times Z e^2}{r_0} $
$\Rightarrow r_0=\frac{9 \times 10^9 \times Z e }{V} $ meter
Putting the values $=9 \times 10^9 \times 1.6 \times 10^{-19}\left(\frac{Z}{V}\right)$ meter
$=14.4 \times 10^{-10}\left(\frac{Z}{V}\right) $ meter
$=\left[14.4\left(\frac{Z}{V}\right)\right] \ \mathring A $

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