A) \[\frac{Z{{e}^{2}}}{2\pi {{\varepsilon }_{0}}m{{v}^{2}}}\]
B) \[\frac{Ze}{4\pi {{\varepsilon }_{0}}m{{v}^{2}}}\]
C) \[\frac{Z{{e}^{2}}}{8\pi {{\varepsilon }_{0}}m{{v}^{2}}}\]
D) \[\frac{Ze}{8\pi {{\varepsilon }_{0}}m{{v}^{2}}}\]
Correct Answer: A
Solution :
Suppose distance of closest approach is r, and according to energy conservation applied for elementary charge. Energy at the time of projection = Energy at the distance of closest approach Þ \[\frac{1}{2}m{{v}^{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{(Ze).e}{r}\Rightarrow r=\frac{Z{{e}^{2}}}{2\pi {{\varepsilon }_{0}}m{{v}^{2}}}\]
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