Question

An alpha nucleus of energy $\frac{1}{2}mv^{2}$ bombards 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/ $Ze$
• D. $1 / v ^{}$

Answer 1

Answer: (B)

At distance of closest approach relative velocity of two particles is $v$ . Here target is considered as stationary, so $\alpha $ -particle comes to rest instantaneously at distance of closest approach. Let required distance is $r,$ then from work energy-theorem.
$0-\frac{m v^{2}}{2}=-\frac{1}{4 \pi \epsilon _{0}}\frac{Z_{e} \times Z_{e}}{r}$
$ \, r \propto \frac{1}{m}$
$ \propto \frac{1}{v^{2}}$
$ \propto Z e^{2}$