AP EAMCET · PHYSICS · Atomic Physics
An alpha particle of energy \(\mathrm{K} \mathrm{MeV}\) is moving towards a nucleus of atomic number \(\mathrm{Z}\). The distance of closest approach of the alpha particle to the nucleus in metres is
- A \(7.2 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{K}}\)
- B \(3.84 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{K}}\)
- C \(14.4 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{K}}\)
- D \(28.8 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{K}}\)
Answer & Solution
Correct Answer
(D) \(28.8 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{K}}\)
Step-by-step Solution
Detailed explanation
At the distance of closest approach the whole kinetic energy get converted into potential energy.…
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