A) v2
B) 1 /m
C) 1/v4
D) 1/Ze
Correct Answer: B
Solution :
Since, here 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 a-particle comes to res-instantaneously at distance of closes- approach. Let required distance is r. then from: \[0-\frac{m{{v}^{2}}}{2}=-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Ze\times 2e}{r}\] \[\Rightarrow \] \[r\propto \frac{1}{m}\] \[\propto \frac{1}{{{v}^{2}}}\] \[\propto Z{{e}^{2}}\]
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