Rutherford discovered the nucleus of the atom by firing alpha particles at gold

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?

Explanation / Answer

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

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