Chapter42 - 1. Kinetic energy (we use the classical formula...

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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 k = 14 0 π ε can be written as 8.99 × 10 9 V·m/C. Thus, from energy conservation, we have ( )( )( ) 91 9 Vm C 12 6 8.99 10 3 1.6 10 C 90 3.00 10 eV e kq q KU r K ×× × =⇒ = = × which yields r = 1.3 × 10 – 13 m (or about 130 fm).
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2. Our calculation is similar to that shown in Sample Problem 42-1. We set ( )( ) 0C u m i n 5.30MeV= 1/ 4 / KU q q r α ε == π and solve for the closest separation, r min : ( )( )( )( ) 19 9 Cu Cu min 6 00 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 qq kqq r KK αα εε ×× === ππ × = 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.
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() ( ) 2 22 Au Au, Au Au, Au 2 Au Au 2 24 11 4 197u 4.00u 5.00MeV 4.00u+197u 0.390MeV. ff i i mm m Km vm v K αα α ⎛⎞ == = ⎜⎟ + + ⎝⎠ = = (b) The final kinetic energy of the alpha particle is Au Au Au Au 2 4.00u 197u 5.00MeV 4.00u 197u 4.61MeV. i i v K −− = ++ = + = We note that K K K af f i += Au, is indeed satisfied. 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 particle is v v fi = + Au Au , and that of the recoiling gold nucleus is v m v Au, Au = + 2 . (a) Therefore, the kinetic energy of the recoiling nucleus is
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4. Using Eq. 42-3 ( 1/3 0 rr A = ), we estimate the nuclear radii of the alpha particle and Al to be 15 15 15 15 Al (1.2 10 m)(4) 1.90 10 m (1.2 10 m)(27) 3.60 10 m. r r α −− = × = × The distance between the centers of the nuclei when their surfaces touch is 15 15 15 Al 1.90 10 m 3.60 10 m 5.50 10 m rr r −− − =+ = × + × = × . From energy conservation, the amount of energy required is 92 2 1 9 1 9 Al 15 0 12 6 1 (8.99 10 N m C )(2 1.6 10 C)(13 1.6 10 C) 45 . 5 0 1 0 m 1.09 10 J 6.79 10 eV qq K r πε ×⋅ × × × × == ×
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which yields 92 2 1 9 1 9 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. qq r K πε −− ×⋅ × × × × == × = 5. 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 KU r
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6. (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.
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This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter42 - 1. Kinetic energy (we use the classical formula...

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