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= 4. An alpha particle, (Q = +2e) with kinetic energy Ki 5.30 MeV happens to be headed directly toward the nucleus of a
= 4. An alpha particle, (Q = +2e) with kinetic energy Ki 5.30 MeV happens to be headed directly toward the nucleus of a neutral gold atom, (QAu = +79e). What is its distance of closest approach d (least center-to-center separation) to the nucleus? Assume that the atom remains stationary.
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Linear Algebra
Authors: Jim Hefferon
1st Edition
978-0982406212, 0982406215
