Question

(b) in the Geiger-Marsden experiment, an a-particle of 6. 12 MeV energy approaches
a gold target (Z = 79), comes momenjarily to rest and then reverses its direction.
Find the distance of closest approach of a particle to the target nucleus
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Answer 1

Answer:

Given,

KE of alpha patricle(k.e.)=6.12Mev

=6.12×10^6×1.6×10^-19J

=9.79×10^-13J

Z=79

D=?

we have

D=1/4pi × 2Ze^2/k.e.

substituting the values

D=2.51×10^-14m