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