Rutherford discovered the nucleus of the atom by firing alpha particles at gold
ID: 1524032 • Letter: R
Question
Rutherford discovered the nucleus of the atom by firing alpha particles at gold foil. Suppose an alpha particle with charge q = +2e and mass m = 6.G4 10^-27 kg is fired directly at a gold nucleus. The gold nucleus has charge of Q = +79e. You may ignore the motion of the gold nucleus in this problem. Suppose an a particles is traveling directly toward a gold nucleus. If the speed of the alpha particle is v = 1.9 10^7 m/s when it is 1 m from the gold nucleus, how close to the gold nucleus will the a particle come before it stops and reverses direction?Explanation / Answer
Ans:-
Use conservation of energy. The speed is low enough that you can get away with classical (non-relativistic) mechanics. We can neglect the impact of the gold's electrons if we end up near the center of their orbitals.
Initial kinetic energy = 1/2 mv^2
Final electrostatic potential energy = k q1 q2 / r
equate and Solve for the distance:
r = 2 k q1 q2 / mv^2
k is coulomb's constant
q1 and q2 are the charges of the alpha and the gold nucleus, which are given in terms of the fundamental charge, e
m is the alpha particle's mass, which is given
v is the speed at which it is fired, also given
r = 2*9*10^9*2*e*79*e/(6.64*10^-27*1.9*10^7)
r = 2.844*10^12*(1.6*10^-19)^2/1.26*10^-19
r =5.78*10^-7m
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