Question

In rutherford's scattering experiment find the distance of closest approach of an alpha particle

Answer 1

Please find below the solution to the asked query:

The distance of closest approach is given as in under:r0 = 14πε0×2×Z×e2KEgiven KE = 6 MeVand assuming the nucleus to be that of copper. then, Z = 29 and e = 1.6×10−19Csubstituting the values in the above equation and calculating we have,r0 = 2 × 10−14 m

Hope this information will clear your doubts about (Atom).

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.