Question

An $\alpha -\text{particle}$ with kinetic energy $10\text{MeV}$ is heading towards a stationary nucleus of atomic number $50$. Calculate the distance of closest approach. Initially they were far apart.

• A. $28.8\times {10}^{-15}\text{m}$
• B. $14.4\times {10}^{-12}\text{m}$
• C. $14.4\times {10}^{-15}\text{m}$
• D. $28.8\times {10}^{-15}\text{m}$

Answer 1

The correct option is C $14.4\times {10}^{-15}\text{m}$

Let the distance of closest approach be $r_{min}$.

At minimum separation the $\alpha$-particle will be at momentary rest.

So, at the distance of closest approach the whole kinetic energy gets converted into potential energy.


$\therefore \text{Decrease in kinetic energy=increase in potential energy}$

$\frac{1}{2}m{v_i}^2-\frac{1}{2}m{v_f}^2=U_f-U_i...\left(1\right)$

Since,
initial kinetic energy, $\frac{1}{2}m{v}_i^2=10\times {10}^6\times 1.6\times {10}^{-19}\text{V}$

initial kinetic energy, $\frac{1}{2}m{v}_f^2=0$

initial potential energy $U_i=0$

final potential energy, $U_f=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r}$

where,
charge of $\alpha -$particle, $q_1=2e=2\times 1.6\times {10}^{-19}\text{C}$,

charge of tin nucleus, $q_2=50e=50\times 1.6\times {10}^{-19}\text{C}$,

$r=r_{min}$

Substituting the values in $\left(1\right)$,

$\Rightarrow \frac{1}{2}m{v}_i^2-0=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r_{min}}-0$

$\therefore r_{min}=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{\frac{1}{2}m{v}_i^2}$

Substituting the values,

$r_{min}=\frac{(9\times {10}^9)(2\times 1.6\times {10}^{-19})(1.6\times {10}^{-19}\times 50)}{10\times {10}^6\times 1.6\times {10}^{-19}}$

$\therefore r_{min}=14.4\times {10}^{-15}\text{m}$

Hence, option (c) is the correct answer.