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Rutherford's atom experiment problem
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[QUOTE="Dana Fishel, post: 4991342, member: 540576"] [h2]Homework Statement [/h2] In 1911, Ernest Rutherford discovered the nucleus of the atom by observing the scattering of helium nuclei from gold nuclei. If a helium nucleus with a mass of 6.68 [ATTACH=full]194772[/ATTACH] 10^-27 kg, a charge of +2[I]e[/I], and an initial velocity of 1.30 [ATTACH=full]194773[/ATTACH] 10^7 m/s is projected head-on toward a gold nucleus with a charge of +79[I]e[/I], how close will the helium atom come to the gold nucleus before it stops and turns around? (Assume the gold nucleus is held in place by other gold atoms and does not move.) [h2]Homework Equations[/h2] electric field equations involving k constant, charge, mass, and velocity [h2]The Attempt at a Solution[/h2] I was told by a classmate that this should work: d = (9e9)4(79)(1.602e-19)^2 / (6.68e-27)(1.3e7) = 8.4e-7 meters (where e means x10^#) But it's wrong, and I'm not sure which equation this solution came from in the first place. [/QUOTE]
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Rutherford's atom experiment problem
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