Question

The distance of closest approach of an $\alpha$-particle fired at nucleus with momentum $P$ is $d$. The distance of closest approach when the $\alpha$-particle is fired at same nucleus with momentum $3P$ will be:

• A. $3d$
• B. $\frac{d}{3}$
• C. $9d$
• D. $\frac{d}{9}$

Answer 1

Answer: (D)

$\frac{d}{9}$

At the distance of closest approach, all of the kinetic energy of the incident particle is converted to potential energy of the system.

Thus $E_i=\frac{P^2}{2m}=\frac{1}{4\pi {ϵ}_0}\frac{Ze\times q}{r}$

where $r$ is the distance of closed approach, $q$ is the charge of incident particle.

Thus for particles with different momenta, $r\propto \frac{1}{P^2}$

Thus for $\alpha$ particle with momentum $3P$, the distance of closest approach becomes $\frac{d}{3^2}=\frac{d}{9}$