Question

Which of the following condition is true for the distance of the closest approach?
1. Potential energy of alpha particle = Kinetic energy of the alpha particle 
2. Potential energy of alpha particle < Kinetic energy of alpha particle 
3. Potential energy of alpha particle > Kinetic energy of the alpha particle 
4. Potential energy  of alpha particle ≠ Kinetic energy of alpha particle 

Answer 1

Correct Answer - Option 1 : Potential energy of alpha particle = Kinetic energy of the alpha particle 

CONCEPT : 

• The distance of the closest approach is the closest distance to the nucleus at which all the kinetic energy of the projected particle becomes potential energy and is given by
  \(\Rightarrow d = \frac{2Ze^{2}}{4\pi \epsilon_{0}K}\)
  Where d = Distance of closest approach, Z = Atomic number, K = Kinetic energy. e = charge

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EXPLANATION :

• The distance of the closest approach is the closest distance to the nucleus at which all the kinetic energy of the projected particle becomes potential energy alpha particle and  target nucleus and is represented by
  \(\frac{1}{2}mV^{2} = \frac{2Ze^{2}}{4\pi\epsilon_{0}d}\)
• Hence option 1 is the answer
• All the other options don't state the condition of kinetic energy is equal to potential energy, which is the required condition for the distance of the closest approach
• Hence,option 2,3 and 4  are incorrect