An -particle collides with nucleus after passing through the pote… - 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 $14.4( Z / V ) \ Å$.
(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)

✓

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.
$\begin{array}{l}
2 e \times V=\frac{1}{4 \pi \in_0}\left(\frac{2 e \times Z e }{r_0}\right)=9 \times 10^9\left(\frac{2 e \times Z e }{r_0}\right) \\
\begin{array}{l}
\Rightarrow 2 eV=\frac{2 \times 9 \times 10^9 \times Z e^2}{r_0} \Rightarrow r_0=\frac{9 \times 10^9 \times Z e }{V} \text { meter } \\
\text { Putting the values }=9 \times 10^9 \times 1.6 \times 10^{-19}\left(\frac{Z}{V}\right) \text { meter } \\
=14.4 \times 10^{-10}\left(\frac{Z}{V}\right) \text { meter } \\
=\left[14.4\left(\frac{Z}{V}\right)\right] \ Å\end{array}\end{array}$

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