Details of Calculating Maximal Nuclear Size
For head on collisions between alpha particles and the nucleus, all the kinetic energy of the alpha particle is turned into potential energy and the particle is at rest. The distance from the centre of the alpha particle to the centre of the nucleus (b) at this point is a maximum value for the radius, if it is evident from the experiment that the particles have not hit the nucleus.
Applying the inverse-square law between the charges on the electron and nucleus, one can write:
Rearranging:
For an alpha particle:
- m (mass) = 6.7×10−27 kg
- q1 = 2×(1.6×10−19) C
- q2 (for gold) = 79×(1.6×10−19) C
- v (initial velocity) = 2×107 m/s
Substituting these in gives the value of about 2.7×10−14 m. (The true radius is about 7.3×10−15 m.) The true radius of the nucleus is not recovered in these experiments because the alphas do not have enough energy to penetrate to more than 27 fm of the nuclear center, as noted, when the actual radius of gold is 7.3 fm. Rutherford realized this, and also realized that actual impact of the alphas on gold causing any force-deviation from that of the 1/r coulomb potential would change the form of his scattering curve at high scattering angles (the smallest impact parameters) from a hyperbola to something else. This was not seen, indicating that the gold had not been "hit" so that Rutherford only knew the gold nucleus (or the sum of the gold and alpha radii) was smaller than 27 fm (2.7×10−14 m)
Read more about this topic: Rutherford Scattering
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