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AP EAMCET202316 May 2023Evening ShiftPhysicsAtomic PhysicsActual

An alpha particle of energy K MeV is moving towards a nucleus of atomic number Z . The distance of closest approach of the alpha particle to the nucleus in metres is

Options

  1. A7.2 10⁻¹⁶ Z K
  2. B3.84 10⁻¹⁶ Z K
  3. C14.4 10⁻¹⁶ Z K
  4. D28.8 10⁻¹⁶ Z K

Correct answer

D. 28.8 10⁻¹⁶ Z K

Step-by-step solution

At the distance of closest approach the whole kinetic energy get converted into potential energy. aligned & K . E = 2 Ze ^2 4 ₀ r & r = 2 ze ^2 4 ₀ k ( mev ) =9 10^9 2 eZ k 10^6 & r =28.8 10⁻¹⁶ Z k aligned

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