Question

An alpha particle of $5MeV$ at a large distance proceeds towards a gold nucleus $(Z=79)$ to make a head on collision. The closest distance of approach from the centre of gold nucleus is:

• A. $20fm$
• B. $15fm$
• C. $10fm$
• D. $45fm$

Answer 1

The correct option is D $45fm$

Initial energy of $\alpha \text{-particle}=V\times e=5\times {10}^6\times (1.6\times {10}^{-19})J.$
Let $x$ be the closest distance with nucleus.
Now closest distance is when the energy of $\alpha$ particle is totally converted to potential energy between them.
$\therefore 5\times {10}^6\times 1.6\times {10}^{-19}=\frac{k{q}_1{q}_2}{r}=\frac{k(79\times e)(2\times e)}{x.}$
$\Rightarrow x=\frac{(9\times {10}^9)(79\times 1\cdot 6\times {10}^{-19})(2\times 1\cdot 6\times {10}^{-19})}{5\times {10}^6\times 1.6\times {10}^{-19}}$
$\Rightarrow x=45\times {10}^{-15}m$