In Rutherford\'s scattering experiments, alpha particles (charge = +2e) were fir

Question

In Rutherford's scattering experiments, alpha particles (charge = +2e) were fired at a gold foil. Consider an alpha particle with an initial kinetic energy K heading directly for the nucleus of a gold atom (charge = +79e). The alpha particle will come to rest when all its initial kinetic energy has been converted to electrical potential energy. Find the distance of closest approach between the alpha particle and the gold nucleus for the case K = 3.6 MeV. Express your answer using two significant figures.

Explanation / Answer

E = qVE = (1/4o)qQ/do    so therefore: do = (1/4o)qQ/E

Using dc = (1/4o)qQ/E we have
dc = [9x109x(2x1.6x10-19x79x1.6x10-19)]/3.6x10-13 = 1.011x10-13 m.
This is well inside the atom but some eight nuclear diameters from the center of the gold nucleus.

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