Question

Rutherford discovered the nucleus of the atom by firing alpha particles at gold foil. An alpha particle has a charge of q=+2e and a ,ass pf m=6.64 x 10^-27 kg. A gp;d nucleus has a charge of Q=+79e. Ignore the motion of the gold nucleus in the problem. Suppose an alpha particle is traveling directly toward a gold nucleus. If the speed of the alpha particle is v=1.9 x 10^7 m/s when it is 1 m from the gold nucleus how close to the gold nucleus will the alpha particle come before it stops and reverses direction?

Answer 1

Given

alpha particle with charge q1 = +2*e = 2*1.6*10^-19 C

and the gold nucleus with charge q2 = +79*e = 79*1.6*10^-19 C

mass of the alpha particle is m = 6.64*10^-27 kg

alpha particle moving with velocity v = 1.9*10^7 m/s

alpha particle is traveling directly toward a gold nucleus by conservation of energy

kinetic energy of the alpha particle = electrostatic potential energy so that the alpha particle momentarily comes to rest and reverses its path

0.5*m*v^2 = k*q1*q2/r

r = k*q1*q2/(0.5*m*v^2)

substituting the values  

r = (9*10^9*2*1.6*10^-19*79*1.6*10^-19)/(0.5*6.64*10^-27*(1.9*10^7)^2)m

r = 3.0373460*10^-14 m

r = 30.37346 fm

the closest distance of the alpha particle before it stops and reversed direction is r = 30.37346 fm