Sunday, February 13, 2011

Mastering Physics: Rutherford Scattering

The first modern models of the atom were formulated by John Dalton, who believed compounds to be made of combinations of atoms. Later J. J. Thomson established a "plum pudding model" of the atom--the first to include electric charge--with negative bits, the "plums," floating in a dispersed amount of positive charge, the "pudding."
In 1908, Ernest Rutherford, along with his students Geiger and Marsden, conducted an experiment to investigate Thomson's model of the atom. They set up an apparatus to have energetic alpha particles (a helium nucleus with two neutrons and two protons) strike a sheet of gold foil. Rutherford and his students expected to see very little deflection of the alpha particles as they passed through the gold foil, because the positive charge would not be densely distributed. However, their experiment showed that an alpha particle would occassionally be reflected at a much greater angle than predicted by the Thomson model. This led Rutherford to surmise the existence of a densely charged positive nucleus at the center of an atom.
By assuming that there was a positive point particle located at the center of the atom, Rutherford was able to develop a formula to predict the percentage of alpha particles that would be scattered from the beam at a particular angle. In this problem, we investigate the scattering of alpha particles by a gold nucleus.

Part A
If an alpha particle (two protons and two neutrons) is given an initial (nonrelativistic) velocity v at a very far distance and is aimed directly at a gold nucleus (Z=79), what is the closest distance d the alpha particle will come to the nucleus? In this problem you can estimate that the mass of the proton m_p is equal to the mass of the neutron and only consider the effects of a single gold nucleus. Assume that the alpha particle comes close enough so that the nucleus is not substantially screened by inner-shell electrons.
Express your answer in terms of m_p, the permittivity of free space epsilon_0, the magnitude of the electron charge e, and v.

  d = \frac{79e^{2}}{4{\pi}{\epsilon}_{0}m_{p}v^{2}}

Part B
The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming alpha particles exceeded 32\; \rm MeV. This can be explained by the fact that the predictions of the formula apply when the only force involved is the electromagnetic force and will break down if the incoming particles make contact with the nucleus. Use the fact that Rutherford's prediction ceases to be valid for alpha particles with an energy greater than 32 \; \rm MeV to estimate the radius r of the gold nucleus.
Express your answer in meters to two significant figures.

  r = 7.10×10−15
 \rm m

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