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hw10_solutions

# hw10_solutions - PHY 251 Fall 2009 homework problem set 10...

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PHY 251 Fall 2009: homework problem set 10, due in the PHY 251 drop box in room A-129 by noon on Friday, Dec. 4. 1. Serway 13.1 Answer: Because of constant nuclear density we have r = r 0 A 1 / 3 with r 0 = 1 . 2 × 10 - 15 m so 4 2 H has r = (1 . 2 × 10 - 15 )(4) 1 / 3 = 1 . 9 × 10 - 15 m 238 92 U has r = (1 . 2 × 10 - 15 )(238) 1 / 3 = 7 . 4 × 10 - 15 m so the ratio is (238 / 4) 1 / 3 = 3 . 9 which is modest for such a large mass ratio of 238 / 4 60 . 2. Serway 13.5 Answer: The turn-around point is when all of the alpha particle’s kinetic energy goes into electrostatic potential energy, or E k = 1 4 πǫ 0 Z α Z r so r = Z α Z 4 πǫ 0 E k = (2)(79) 4 π (8 . 85 × 10 - 12 ) · (0 . 5 × 10 6 eV ) · (1 . 602 × 10 - 19 J/eV ) = 4 . 5 × 10 - 13 m or 450 fm. To approach within 300 fm we must have a kinetic energy of E k = 1 4 πǫ 0 (2)(79)(1 . 602 × 10 - 19 Coulomb ) 2 300 × 10 - 15 = 1 . 21 × 10 - 13 Joules or if we divide by 1 . 602 × 10 - 19 J/eV an energy of 0.75 MeV. This energy is small compared to the alpha particle rest mass of about 4 · 940 MeV so we can find the velocity using a non-relativistic formula: v = radicalBigg 2 E k m = radicalBigg 2 E k mc 2 · c = radicalBigg 2 · 0 . 75 4 · 940 · 2 . 99 × 10 8 = 6 . 0 × 10 6 m/s . 3. Serway 13.9 Answer: We assume that the net atomic weight of copper m is due to a fraction x times the weight m 63 of 63 Cu and a fraction ( 1 x ) of the weight m 65 of 65 Cu. That is, m = xm 63 + (1 x ) m 65 m m 65 = x ( m 63 m 65 ) x = m m 65 m 63 m 65 = 63 . 55 64 . 95 62 . 95 64 . 95 = 0 . 70 so we have 70% 63 Cu and 30% 65 Cu. 4. Serway 13.16. Additional places to get atomic masses are at http://physics.nist.gov/PhysRefData/Compositions/index.html and http://www.nndc.bnl.gov/nudat2/ Answer: The desired reaction is 43 20 Ca 23 42 20 Ca 22 + 1 0 n 1 . Using Appendix B, we can get 1

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