In a head on collision between an α -particle and gold nucleus, the closest distance of approach is 41.3 fm. Calculate the energy of α -particle (Z of gold =79)

Question

In a head on collision between an α -particle and gold nucleus, the closest distance of approach is 41.3 fm. Calculate the energy of α -particle (Z of gold =79) (A) 5.51 eV (B) 5.51 MeV (C) 5.51 J (D) 5.51 KJ

Tardigrade Admin · Accepted Answer

E =(( Ze )(2 e )/4 π ε0 r )=(9 × 109 × 79 × 2(1.6 × 10-19)2/41.3 × 10-15) =(9 × 79 × 3.2 × 1.6 × 10-14/41.3) J =8.814 × 10-13 Joule =(8.814 × 10-13/1.6 × 10-13) E =5.51 MeV

Q. In a head on collision between an α -particle and gold nucleus, the closest distance of approach is 41.3fm. Calculate the energy of α -particle (Z of gold =79)

5.51 eV

5.51 MeV

5.51 J

5.51 KJ

E=4πε0​r(Ze)(2e)​=41.3×10−159×109×79×2(1.6×10−19)2​ =41.39×79×3.2×1.6×10−14​J =8.814×10−13 Joule =1.6×10−138.814×10−13​ E=5.51MeV

Q. In a head on collision between an $α$ -particle and gold nucleus, the closest distance of approach is $41.3 f m$ . Calculate the energy of $α$ -particle $(Z$ of gold $= 79)$