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)
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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