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Distance of closest approach
Concepts tested here
- closest-approach-mass-dependence
All Questions
2016 1 question
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When an $\alpha$-particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as
$r_0 = \frac{4Ze^2}{4\pi\varepsilon_0mv^2}$.
By conservation of energy, the initial KE of the $\alpha$-particle equals the potential energy at closest approach:
$\frac{1}{2}mv^2 = \frac{2Ze^2}{4\pi\varepsilon_0r_0}$
So $r_0 \propto \frac{1}{m}$