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An -particle of energy 5 MeV is scattered through 180^ by a fixed uranium nucleus. The distance of the closest approach is of the order of

Options

  1. A1 Å
  2. B10⁻¹⁰ ~cm
  3. C10⁻¹² ~cm
  4. D10⁻¹⁵ ~cm

Correct answer

C. 10⁻¹² ~cm

Step-by-step solution

At closest approach, all the kinetic energy of the -particle will converted into the potential energy of the system, K.E. = P.E. aligned & 5 MeV = 1 4 ₀ q ₁ q ₂ r & 5 10^6 e =9 10^9 Z ₁ Z ₂ e ^2 r & r = 9 10^9 92 2 1.6 10⁻¹⁹ 5 10^6 & r =5.3 10⁻¹⁴ ~m =5.3 10⁻¹² ~cm aligned

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