Question

In Rutherford's scattering experiments, alpha particles (charge = +2e) were fired at a gold foil. Consider an alpha particle, very far from the gold foil, with an initial kinetic energy of 2.0 MeV heading directly for a gold atom (charge +79e). The alpha particle will come to rest when all its initial kinetic energy has been converted to electrical potential energy. Find the distance of closest approach between the alpha particle and the gold nucleus.

Answer 1

convert 2MeV to joules.
energy=2eVE6=2E6*1.602×10−13 J=2*1.6E-7 Joules

Now set that energy to the PE at the closest point

EPE=k2e*79e/d=8.98*2*(1.6^2)(79)(1E-26)/r

set it equal to the initial KE
2*1.6E-7*d=8.98*2*1.6^2 E-26*79)

d= 8.98*79*1.6 E -19 meters

check that.

Answer 2

ÔÚ