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Q.
If an $ \alpha $ -particle of mass m, charge q and velocity v is incident on a nucleus of charge Q and mass m, then the distance of closest approach is
VMMC MedicalVMMC Medical 2009
Solution:
The minimum distance from the nucleus up to which the $ \alpha - $ particle approach, is called the distance of closest approach fro). At this distance the entire initial kinetic energy is converted into potential energy. So, $ \frac{1}{2}m{{v}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{r} $ $ \Rightarrow $ $ r=\frac{qQ}{2\pi {{\varepsilon }_{0}}m{{v}^{2}}} $ None of the given option is correct.