Question

An alpha nucleus of energy $\frac{1}{2}mv^{2}$ bombards on a heavy nuclear target of the charge $\textit{Ze}$ . Then the distance of closest approach for the alpha nucleus will be proportional to

• A. $\frac{Z^{2}}{v^{2}}$
• B. $\frac{v^{2}}{Z^{2}}$
• C. $\frac{Z}{v^{2}}$
• D. $\frac{v^{2}}{Z}$

Answer 1

Answer: (C)

$\frac{Z}{v^{2}}$

Answer 2

For the closest approach, kinetic energy is converted into potential energy $\frac{1}{2} m v^{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{0}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(Z e)(2 e)}{r_{0}}$ $r_{0} \propto\left(\frac{Z}{v^{2}}\right)$