A) \[\frac{1}{Ze}\]
B) \[{{V}^{2}}\]
C) \[\frac{1}{m}\]
D) \[\frac{1}{{{V}^{4}}}\]
Correct Answer: C
Solution :
An \[\alpha \text{-}\]particle of mass m passes initial velocity v, when it is at a large distance from the nucleus of an atom having atomic number Z. At the distance of closer approach, the kinetic energy of \[\alpha \text{-}\]particle is completely converted into potential energy. \[\frac{1}{2}m{{v}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(Ze)\,(Ze)}{{{r}_{0}}}\] \[{{r}_{0}}=\frac{1}{4\pi {{\varepsilon }_{0}}},\frac{2Z{{e}^{2}}}{\frac{1}{2}m{{v}^{2}}}\]
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