CUET · PHYSICS · PYQ PAPER 2025
When an \( \alpha \)-particle of mass M moving with velocity v bombards a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on M as
- A Independent of M
- B \( \propto \frac{1}{M^2} \)
- C \( \propto \frac{1}{M} \)
- D \( \propto \frac{1}{\sqrt{M}} \)
Answer & Solution
Correct Answer
(C) \( \propto \frac{1}{M} \)
Step-by-step Solution
Detailed explanation
Kinetic Energy = Potential Energy at closest approach: \( \frac{1}{2} M v^2 = \frac{1}{4 \pi \epsilon_0} \frac{(2e)(Ze)}{r_0} \) Solve for \( r_0 \): \( r_0 = \frac{1}{4 \pi \epsilon_0} \frac{4Ze^2}{M v^2} \) Therefore, \( r_0 \propto \frac{1}{M} \)
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