An alpha-particle is projected towards a stationary Cu nucleus, an Au nucleus and an Ag nucleus with same speed from a large distance. The distances of closest approaches were found to be $d_{1}$ , $d_{2}$ and $d_{3}$ respectively. Which of the following is correct?

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$d_{1} < d_{3} < d_{2}$

$d_{1} < d_{2} < d_{3}$

$d_{2} < d_{3} < d_{1}$

$d_{2} < d_{1} < d_{3}$

answer is C.

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Detailed Solution

$f o r d i s \tan c e o f c l o s e s t a p p r o a c h \frac{1}{2} m v^{2} = \frac{Z e 2 e}{4 π ε_{0} d} d i s p r o p o r t i o n a l t o Z Z f o r A g 47 f o r A u 79 f o r C u 29$

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