Question

An $\alpha -$particle of energy $3\text{MeV}$ is scattered through ${180}^{\circ }$ by gold nucleus. The distance of the closest approach is of the order of -

The atomic number of gold is $79$.

• A. ${10}^{-11}\text{cm}$
• B. ${10}^{-19}\text{cm}$
• C. ${10}^{-13}\text{cm}$
• D. ${10}^{-16}\text{cm}$

Answer 1

The correct option is A ${10}^{-11}\text{cm}$

From law of conservation of mechanical energy,

$K{E}_i+P{E}_i=K{E}_f+P{E}_f$

$\Rightarrow 3\times {10}^6\times 1.6\times {10}^{-19}+0=0+\frac{9\times {10}^9\times (79\times 1.6\times {10}^{-19})\times (2\times 1.6\times {10}^{-19})}{r}$

$\Rightarrow r=7.6\times {10}^{-14}\text{m}=7.6\times {10}^{-12}\text{cm}\approx {10}^{-11}\text{cm}$

Therefore, the distance of the closest approach is of the order of ${10}^{-11}\text{cm}$.

Hence, option $\left(A\right)$ is the correct answer.