1179_Physics ProblemsTechnical Physics

1179_Physics ProblemsTechnical Physics - 520 P42.4 Atomic...

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520 Atomic Physics P42.4 (a) The point of closest approach is found when EKU kqq r e =+=+ 0 α Au or r ke e E e min = 27 9 a fa f r min .. . = ×⋅ × × 8 9 91 0 1 5 81 6 01 0 4 00 1 60 10 568 10 9 2 13 14 Nm C C MeV J MeV m 22 ej af . (b) The maximum force exerted on the alpha particle is F kq q r e max min . . == × × = Au C m N 2 9 2 14 2 8 9 0 1 5 6 0 11 3 away from the nucleus. Section 42.3 Bohr’s Model of the Hydrogen Atom P42.5 (a) v ke mr e e 1 2 1 = where ra 1 2 0 11 1 0 005 29 5 29 10 = × a f nm m v 1 9 2 31 11 6 8 99 10 1 60 10 9 11 10 5 29 10 219 10 = × ×× −− . C kg m ms e j (b) Km v e 11 23 1 6 2 18 1 2 1 2 9 11 10 2 19 10 2 18 10 13 6 × × = × = . . kg m s J eV (c) U r e 1 2 1 9 2 11 18 8 99 10 1 60 10 529 10 435 10 272 =− × × × = − . C m J e V e j P42.6 E nn if F H G I K J 13 6 . eV a f Where for E > 0 we have absorption and for E < 0 we have emission. (i) for n i = 2 and n f = 5, E = 286 . eV (absorption) (ii) for n i =
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