3)alpha particle of 5.0MeV are being scattered by a thin copper(Z=29)foil.up to what closet dustance can these particles reach the nucleus of copper ?if there be protons of same energy ,then?
Ans is 1.67×10^-14m,0.835×10^-14m
The formula for distance (r) = Ze×2e/4π€×KE
GIVEN
Z=29
KE= 5×1.6×10^-13
r= 9*10^9 ×29×1.6×10^-13×2×1.6×10^-13/
5×1.6×10^-13
r=167.04×10^-12
r =1.6×10^-14