Q10 of 26 Page 1

Determine the distance of closest approach when an alpha particle of kinetic energy 4 �5 MeV strikes a nucleus of Z = 80, stops and reverses its direction.

Charge on the alpha particle, Q1 = 2e = 2 × 1.6 × 10-19 C = 3.2 × 10-19 C

Charge on the Z=80 nucleus, Q2 = 80e = 80 × 1.6 × 10-19 C = 128 × 10-19 C


The energy with which the alpha particle is incident on the nucleus,


E0 = 4.5 MeV


= 4500 eV


= 4500 × 1.6 ×10-19 J


= 7.2 × 10-13 J





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