ch42 - Chapter 42 1. Kinetic energy (we use the classical...

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1565 Chapter 42 1. Kinetic energy (we use the classical formula since v is much less than c ) is converted into potential energy (see Eq. 24-43). From Appendix F or G, we find Z = 3 for lithium and Z = 90 for thorium; the charges on those nuclei are therefore 3 e and 90 e , respectively. We manipulate the terms so that one of the factors of e cancels the “e” in the kinetic energy unit MeV, and the other factor of e is set to be 1.6 10 –19 C. We note that 0 14 k  can be written as 8.99 10 9 V·m/C. Thus, from energy conservation, we have     9 19 V m C 1 2 6 8.99 10 3 1.6 10 C 90 3.00 10 eV e kq q K U r K which yields r = 1.3 10 – 13 m (or about 130 fm). 2. Our calculation is similar to that shown in Sample Problem — “Rutherford scattering of an alpha particle by a gold nucleus.” We set K 4.60MeV= U 1/ 4 0 q q Cu / r min and solve for the closest separation, r min : r min q q Cu 4  0 K kq q Cu 4  0 K 2 e   29   1.60 10 19 C 8.99 10 9 V m/C 5.30 10 6 eV 1.81 10 14 m 18.1 fm. We note that the factor of e in q = 2 e was not set equal to 1.60 10 – 19 C, but was instead allowed to cancel the “e” in the non-SI energy unit, electron-volt. 3. Kinetic energy (we use the classical formula since v is much less than c ) is converted into potential energy. From Appendix F or G, we find Z = 3 for lithium and Z = 110 for Ds; the charges on those nuclei are therefore 3 e and 110 e , respectively. From energy conservation, we have Li Ds 0 1 4 q q K U r which yields
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CHAPTER 42 1566 9 2 2 19 19 Li Ds 13 0 14 1 (8.99 10 N m C )(3 1.6 10 C)(110 1.6 10 C) 4 (10.2 MeV)(1.60 10 J/MeV) 4.65 10 m 46.5 fm. q q r K  4. In order for the particle to penetrate the gold nucleus, the separation between the centers of mass of the two particles must be no greater than r = r Cu + r = 6.23 fm + 1.80 fm = 8.03 fm. Thus, the minimum energy K is given by      Au Au 0 9 19 6 15 1 4 8.99 10 V m/C 2 79 1.60 10 C 28.3 10 eV. 8.03 10 m q q kq q K U r r e We note that the factor of e in q = 2 e was not set equal to 1.60 10 – 19 C, but was instead carried through to become part of the final units. 5. The conservation laws of (classical kinetic) energy and (linear) momentum determine the outcome of the collision (see Chapter 9). The final speed of the particle is v m m m m v f i Au Au , and that of the recoiling gold nucleus is v m m m v f i Au, Au 2 . (a) Therefore, the kinetic energy of the recoiling nucleus is
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ch42 - Chapter 42 1. Kinetic energy (we use the classical...

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