A) \[\frac{Qq}{4\pi {{\varepsilon }_{0}}{{m}^{2}}{{v}^{2}}}\]
B) \[\frac{Qq}{4\pi {{\varepsilon }_{0}}m{{v}^{2}}}\]
C) \[\frac{Qqm{{v}^{2}}}{2}\]
D) \[\frac{Qq}{m{{v}^{2}}}\]
Correct Answer: B
Solution :
The minimum distance from the nucleus up to which the a-particle approach, is called the distance of closest approach \[({{r}_{0}})\]. 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.
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