AP EAMCET · PHYSICS · Atomic Physics
The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the nucleus with linear momentum \(\mathrm{P}\) is \(\mathrm{d}\). The distance of closest approach of alpha particle to nucleus, if the linear momentum of the alpha particle is \(1.5 \mathrm{P}\)
- A \(\frac{2 \mathrm{~d}}{3}\)
- B \(\frac{3 \mathrm{~d}}{2}\)
- C \(\frac{4 d}{9}\)
- D \(\frac{9 \mathrm{~d}}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{4 d}{9}\)
Step-by-step Solution
Detailed explanation
At sufficient distance, electric potential energy is zero. So, \(\mathrm{U}_1=0\) at (1) \(\mathrm{K}_1=\frac{1}{2} \mathrm{mv}^2=\frac{\mathrm{P}^2}{2 \mathrm{~m}}\) at (2) \(\mathrm{K}_2=0\)…
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