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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

  1. A \(7.2 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{K}}\)
  2. B \(3.84 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{K}}\)
  3. C \(14.4 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{K}}\)
  4. D \(28.8 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{K}}\)
Verified Solution

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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