Question

An alpha nucleus of energy $\frac{1}{2}m{v}^2$ is bombarded onto a heavy nuclear target of charge $Ze$. Then the distance of closest approach for the alpha nucleus will be proportional to

• A. $v^2$
• B. $1/m$
• C. $1/v^4$
• D. $1/Ze$

Answer 1

The correct option is B $1/m$

Kinetic energy of alpha particle is $=\frac{1}{2}m{v}^2$

Potential energy = $\frac{1}{4\pi {\epsilon }_0}\frac{q{z}_e}{r}$

According to conservation of energy

KE = PE

$\frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\frac{q{z}_e}{r}$

$r=\frac{1}{\pi {\epsilon }_0}\frac{z{e}^2}{m{v}^2}$

$r\propto \frac{1}{m}$