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Unformatted text preview: 1. Our calculation is similar to that shown in Sample Problem 421. We set ( ) ( ) C u m i n 5.30MeV= 1/ 4 / K U q q r α ε = = π and solve for the closest separation, r min : ( )( ) ( ) ( ) 19 9 Cu Cu min 6 14 2 29 1.60 10 C 8.99 10 V m/C 4 4 5.30 10 eV 1.58 10 m 15.8 fm. e q q kq q r K K α α ε ε − − × × ⋅ = = = π π × = × = We note that the factor of e in q α = 2 e was not set equal to 1.60 × 10 – 19 C, but was instead allowed to cancel the “e” in the nonSI energy unit, electronvolt. 2. Kinetic energy (we use the classical formula since v is much less than c ) is converted into potential energy (see Eq. 2443). 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 equal to its SI value 1.6 × 10 –19 C. We note that k = 1 4 π ε can be written as 8.99 × 10 9 V·m/C. Thus, from energy conservation, we have K U r k K e q q = ¡ = = × × × × ⋅ − 1 2 9 1 9 6 8 99 10 3 16 10 90 300 10 . . . V m C C eV c hc hb g which yields r = 1.3 × 10 – 13 m (or about 130 fm). 3. The conservation laws of (classical kinetic) energy and (linear) momentum determine the outcome of the collision (see Chapter 9). The final speed of the α particle is v m m m m v f i α α α α = − + Au Au , and that of the recoiling gold nucleus is v m m m v f i Au, Au = + 2 α α α . (a) Therefore, the kinetic energy of the recoiling nucleus is ( ) ( ) ( ) ( ) ( ) 2 2 2 Au Au, Au Au, Au 2 Au Au 2 2 4 1 1 2 2 4 197u 4.00u 5.00MeV 4.00u+197u 0.390MeV. f f i i m m m K m v m v K m m m m α α α α α α § · = = = ¨ ¸ + + © ¹ = = (b) The final kinetic energy of the alpha particle is ( ) 2 2 2 2 Au Au Au Au 2 1 1 2 2 4.00u 197u 5.00MeV 4.00u 197u 4.61MeV. f f i i m m m m K m v m v K m m m m α α α α α α α α α α § · § · − − = = = ¨ ¸ ¨ ¸ + + © ¹ © ¹ § · − = ¨ ¸ + © ¹ = We note that K K K af f i + = Au, α is indeed satisfied. 4. (a) 6 protons, since Z = 6 for carbon (see Appendix F). (b) 8 neutrons, since A – Z = 14 – 6 = 8 (see Eq. 421). 5. (a) Table 421 gives the atomic mass of 1 H as m = 1.007825 u. Therefore, the mass excess for 1 H is ∆ = (1.007825 u – 1.000000 u)= 0.007825 u. (b) In the unit MeV/ c 2 , ∆ = (1.007825 u – 1.000000 u)(931.5 MeV/ c 2 ·u) = +7.290 MeV/ c 2 . (c) The mass of the neutron is given in Sample Problem 423. Thus, for the neutron, ∆ = (1.008665 u – 1.000000 u) = 0.008665 u. (d) In the unit MeV/ c 2 , ∆ = (1.008665 u – 1.000000 u)(931.5 MeV/ c 2 ·u) = +8.071 MeV/ c 2 ....
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This note was uploaded on 10/01/2007 for the course PHYS 2213 taught by Professor Perelstein,m during the Fall '07 term at Cornell University (Engineering School).
 Fall '07
 PERELSTEIN,M
 Atom, Magnetism, Kinetic Energy, Mass, Nuclear Fission, Neutron, Heat, Binding energy

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