Question

An $\alpha -particle$ of energy $5MeV$ is scattered through ${180}^{\circ }$ by gold nucleus. The distance of closest approach is of the order of

• A. ${10}^{-10}cm$
• B. ${10}^{-12}cm$
• C. ${10}^{-14}cm$
• D. ${10}^{-16}cm$

Answer 1

The correct option is B ${10}^{-12}cm$

Atomic number of gold $Z=79$

Kinetic energy of $\alpha$ particle $K.E=5MeV$

At closest approach, all the kinetic energy of the $\alpha$ particle has converted into its potential energy i.e. K.E = P.E

Potential energy at closest approach $P.E=\frac{kZe\left(2e\right)}{r}$

But $\frac{2kZ{e}^2}{r}=5$ $MeV$

$\therefore$ $\frac{2(9\times {10}^9)\times 79\times (1.6\times {10}^{-19}{)}^2}{r}=5$ $\times 1.6\times {10}^{-13}$ $⟹r\approx 4.55\times {10}^{-14}$ m

Hence closest approach is of the order of ${10}^{-14}$ $m$ OR ${10}^{-12}$ $cm$