Physics, asked by sabbahkhan3952, 1 year ago

A beam of a particle velocity 2.110^7m/s is scattered by a gold foil(z=79). Find the distance of closest approach of the a- particle to the gold nucleus the value of charge /mass for a-particle is 4.810^7c/kg find

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Answered by udayjogi

12

concept of distance of closest approach

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