An alpha particle of energy 5 MeV is directed towards a lead nucleus. Find the distance of closest approach

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
  
  Where d = Distance of closest approach, Z = Atomic number, K = Kinetic energy. e = charge 

CALCULATION : 

Given - K = 5 meV = 5 × 1.6 × 10-13J, Z = 82, and e = 1.6 × 10-19 C

• The distance of closest approach is given by
  

Substituting the given values the above equation becomes 

• Hence, option 2 is the answer