an alfa particle of energy 4 Mev is scattered by gold foil (Z = 79) calculate the maximum - Physics - Atoms

Question

an alfa particle of energy 4 Mev is scattered by gold foil (Z = 79)
calculate the maximum volume in which positive charge of the atom
is likely to be concentrated

Ravi Gupta · Accepted Answer

rmin = back limit scattering rmin = Zke2KE=79×9×109×1
6×1 6×10-384×106×1 6×10-19= 284
4×10-16 mvolume of the sphere V = 43πrmin3 = 43×3
14×284 4×10-163= 0 9×10-40m3

an alfa particle of energy 4 Mev is scattered by gold foil. (Z = 79). calculate the maximum volume in which positive charge of the atom is likely to be concentrated.