# Model assume the 238 u nucleus is at rest energy is

This preview shows pages 42–47. Sign up to view the full content.

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 = = ×

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 = = × × = × =
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 = = × ×

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 × + = × × × × = × = × × × =
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern