A) \[{{v}^{2}}\]
B) \[1/m\]
C) \[1/{{v}^{4}}\]
D) \[1/Ze\]
Correct Answer: B
Solution :
At the distance of closest approach, the total kinetic energy of particle will must be converted into potential energy. Since, nuclear target is heavy, it can be assumed safely that it will remain stationary and will not move due to the coulombic interaction force. At distance of closest approach, relative velocity of two particles is v. Here, target is considered as stationary, so \[\alpha \]-particle comes to rest instantaneously at distance of closest approach. Let ' required distance be r, then from work-energy theorem, \[0-\frac{m{{v}^{2}}}{2}=-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Ze\times 2e}{r}\] \[(\because for\,He,q=Ze)\] \[\Rightarrow \] \[r\propto \frac{1}{m}\]or \[\propto \frac{1}{{{v}^{2}}}\propto Z{{e}^{2}}\]
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