Question: = 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

= 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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