Rankers Physics

Atomic Structure: Practice Problem & Solution

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: (2010 Pre)
$1/v^4$
$1/Ze$
$v^2$
$1/m$

Solution Explained:

To solve this problem, we apply the core principles of Atomic Structure. Understanding the underlying formula is key to arriving at the correct answer below:

The distance of closest approach is found by equating kinetic energy to electrostatic potential energy: $K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0}$. Since $K = \frac{1}{2} m v^2$, we have $r_0 \propto \frac{1}{m}$.

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