Question: Q. 12. An $\alpha$-particle moving with initial kinetic energy $K$ towards a nucleus of atomic number $Z$ approaches a distance ’ $d$ ’ at which it reverses its direction. Obtain the expression for the distance of closest approach ’ $d$ ’ in terms of the kinetic energy of $\alpha$-particle $K$.
A [OD 2014]
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Solution:
Ans. At the distance, $d$, the K.E. (K) gets converted into P.E. of the system.
Hence $\quad K=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 e \mathrm{Ze}}{d}$
$$ d=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 e Z e}{K} $$
