Posted by mdave on .
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 3.3 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.
M V^2/2 = Z*2*e^2/d
where M is the mass of the alpha particle and Z is the atomic number of gold (79). Solve for the minimum separation, d.
V is the velocity associated with the 3.3 MeV energy of the alpha particle.
(Or just convert 3.3 MeV to Joules for the left side)
Thank you, but in the above equation what does the e in e^2 equal?