Question

An alpha particle was projected with 5.4 mev energy toward nucleus of cu z(29) what is the closest distance of approach.

Answer 1

Answer :

9.6×10−15m

Solution :

Here, E=8.7MeV=8.7×1.6×10−13J

Z=29,r0=?

r0=2Ze24π∈0(E)

=9×109×2×29(1.6×10−19)28.7×1.6×10−13

r0=9.6×10−15m