Question

An alpha nucleus of energy $\overline{2}$ mv bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

• A. $\frac{1}{Ze}$
• B. $v^2$
• C. $\frac{1}{m}$
• D. $\frac{1}{v^4}$

Answer 1

The correct option is C $\frac{1}{m}$

Let $r$ be the distance of closest approach.

At the closest approach, all the kinetic energy of alpha nucleus has converted into the potential energy i.e. $K.E=P.E$

$\therefore$ $\frac{1}{2}m{v}^2=\frac{kZ{e}^2}{r}$

We get, $r=\frac{2kZ{e}^2}{m{v}^2}$

$⟹r\propto \frac{1}{m}$