Model assume the 238 u nucleus is at rest energy is

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Model: Assume the 238 U nucleus is at rest. Energy is conserved. Visualize: Solve: The energy conservation equation K f + U f = K i + U i is i 15 0 9 2 2 19 2 12 i 15 19 1 (2 )(92 ) 0 J 0 4 7.5 10 m (9.0 10 N m /C )184(1.60 10 C) 1 eV 5.65 10 J 35.33 MeV 7.5 10 m 1.60 10 J e e K K πε + = + × × × = = × × = × × That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e gains kinetic energy K eV by accelerating through a potential difference V = 1 2 K V. Thus the alpha particle will just reach the 238 U nucleus after being accelerated through a potential difference 7 35.33 MV 1.8 10 V 2 V Δ = = ×
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38.43. Model: Assume the 16 O nucleus is at rest. Energy is conserved. Visualize: A proton with initial velocity v i and zero electric potential energy (due to its infinite separation from an 16 O nucleus) moves toward 16 O and turns around when it is 1.0 fm from the surface. Since r is measured from the center of the nucleus, f 4 fm. r = Solve: (a) The energy conservation equation K f + U f = K i + U i is ( ) ( ) 2 1 i 2 0 8 1 0 J 0 J 4 3.0 fm 1.0 fm e e mv πε + = + + ( ) ( ) ( ) ( ) ( )( ) 2 9 2 2 19 2 14 2 2 7 i i 15 27 2 9.0 10 N m /C 8 1.6 10 C 5.52 10 m /s 2.3 10 m/s 4.0 10 m 1.67 10 kg v v × × = = × = × × × (b) ( )( ) 2 27 14 2 2 13 1 1 i 2 2 1.67 10 kg 5.52 10 m /s 4.60 10 J 2.9 MeV K mv = = × × = × =
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38.44. Model: Assume the 12 C nucleus is at rest. Energy is conserved. Visualize: Solve: (a) A proton with an initial velocity v i and zero electric potential energy (due to its infinite separation from a 12 C nucleus) moves toward the 12 C nucleus. It must impact the 12 C nucleus, which has a radius of 2.75 fm, with an energy of 3.0 MeV. The energy conservation equation K f + U f = K i + U i is ( ) ( ) 2 proton i 15 0 6 1 1 3.0 MeV 0 J 4 2 2.75 10 m e e m v πε + = + × ( ) ( ) ( ) ( ) 2 9 2 2 19 19 6 27 2 i 15 9.0 10 N m /C 6 1.60 10 C 1.6 10 J 1 3.0 10 eV 1.67 10 kg 1 eV 2.75 10 m 2 v × × × × × + = × × 7 i 3.43 10 m/s v = × (b) The initial kinetic energy is 2 i i 1 . 2 K mv e V = = Δ Hence, ( )( ) 2 27 7 6 19 1.67 10 kg 3.43 10 m/s 1 6.14 10 V 2 1.6 10 C V × × Δ = = × ×
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38.45. Model: Assume the 137 Cs nucleus is at rest. Energy is conserved. Visualize: The initial kinetic energy of the beta particle as it is ejected from the 137 Cs nucleus is K i . Because the beta particle is ejected from the surface, it also has electric potential energy U i from the coulombic attraction to the nucleus. As the beta particle moves away from the nucleus and is practically at an infinite distance, U f = 0 J and K f = 300 keV. Solve: The conservation of energy equation K f + U f = K i + U i is ( ) ( ) ( ) ( ) ( ) 3 i 15 0 2 9 2 2 19 5 5 i i 15 19 i 55 1 300 10 eV 0 J 4 6.2 10 m 9.0 10 N m /C 55 1.6 10 C 1 eV 3.0 10 eV 127.6 10 eV 6.2 10 m 1.6 10 J 13.1 MeV e e K K K K πε × + = × × × × = × = × × × =
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38.46. Model: Momentum is conserved during the interaction. Assume the nucleus is initially at rest. Visualize: Solve: Momentum is a vector, so both the x - and y -components of G p must be conserved. The x -equation is 7 7 i f n fn fn 5 fn fn 4(1.50 10 m/s) cos cos 4(1.49 10 m/s) cos49 197 cos cos ( ) 1.0609 10 m/s x m v m v m v v v v α α α α θ φ φ φ = × = + = × ° + = = × The y -equation is 7 f n fn fn 5 fn fn 0 sin sin 4(1.49 10 m/s) sin49 197 sin sin ( ) 2.283 10 m/s y
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