Ch42 - 1 Our calculation is similar to that shown in Sample Problem 42-1 We set C u m i n 5.30MeV= 1 4 K U q q r α ε = = π and solve for the

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Unformatted text preview: 1. Our calculation is similar to that shown in Sample Problem 42-1. 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 non-SI 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. 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 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. 42-1). 5. (a) Table 42-1 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 42-3. 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).

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Ch42 - 1 Our calculation is similar to that shown in Sample Problem 42-1 We set C u m i n 5.30MeV= 1 4 K U q q r α ε = = π and solve for the

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