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Q.
An $\alpha$ -particle of $10 \,MeV$ collides head on with a copper nucleus $(Z=29)$ and is deflected back. Then the minimum distance of approach between the centres of the two is
Atoms
Solution:
Let the minimum distance of approach be $r_{0}$. At this distance, the whole of the kinetic energy of the alpha-particle will be converted into the electrical potential energy.
The positive charge on the copper nucleus $=Z e=29 e$ and the positive charge on the $\alpha$ -particle $=2 e$
At $r=r_{0}, \,\,\, KE = PE$
$10 \times\left(1.6 \times 10^{-13}\right)=\frac{1}{4 \pi \varepsilon_{0}} \frac{(29 e)(2 e)}{r_{0}}$
$\therefore r_{0}=\frac{\left(9 \times 10^{9}\right)(29)(2)\left(1.6 \times 10^{-19}\right)^{2}}{10 \times\left(1.6 \times 10^{-13}\right)}=8.4 \times 10^{-15} \,m$