An alpha particle of energy \(\dfrac{1}{2}\) mv\(^2\) bombards a heavy target nucleus of charge Ze. Then, the distance of closet approach for the alpha particle will be proportional to
Show Hint
At closest approach, kinetic energy equals electric potential energy, so \(r_0 \propto 1/v^2\).
Step 1: Understand the concept:
In head-on approach, the alpha particle slows down because the nucleus repels it. At the closest point, its speed becomes zero and all its kinetic energy has changed into electric potential energy.
Step 2: Write the energy balance:
The alpha particle has charge \(2e\). The potential energy at distance \(r\) from a nucleus of charge \(Ze\) is \(\dfrac{1}{4\pi\epsilon_0}\dfrac{(2e)(Ze)}{r}\). So \[ \frac{1}{2} m v^2 = \frac{1}{4\pi\epsilon_0}\frac{2Ze^2}{r_0} \]
Step 3: Solve for r0:
\[ r_0 = \frac{1}{4\pi\epsilon_0}\frac{2Ze^2}{\frac{1}{2}mv^2} = \frac{Ze^2}{\pi\epsilon_0 m v^2} \] So \(r_0 \propto \dfrac{Ze^2}{m v^2}\).
Step 4: Read the proportionality:
The distance goes as \(v^{-2}\) and grows directly with \(Z\). Option (1) is wrong because the dependence on speed is squared. Option (3) has too high a power. Option (4) is wrong because \(r_0\) is proportional to \(Ze\), not inversely proportional to it.
Final Answer:
The distance of closest approach is proportional to \(v^{-2}\), option (2). \[ \boxed{r_0 \propto v^{-2}} \]
