An alpha particle with kinetic energy 11.0 MeVmakes a head-on collision with a l

Question

An alpha particle with kinetic energy 11.0 MeVmakes a head-on collision with a lead nucleus at rest.

Part A

What is the distance of closest approach of the two particles? (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.)

An alpha particle with kinetic energy 11.0 MeVmakes a head-on collision with a lead nucleus at rest.

Part A

What is the distance of closest approach of the two particles? (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.)

Explanation / Answer

rmin=kq1q2/KE=9x109 x 2 x1.6x10-19 x82x1.6x10-19/11x106x1.6x10-19=214.69x10-16m=21.5fm

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