{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1274_Physics ProblemsTechnical Physics

1274_Physics ProblemsTechnical Physics - Chapter 45...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 45 615 P45.58 (a) The mass of the pellet is m V = = × F H G I K J L N M M O Q P P = × ρ π 0 200 4 3 1 50 10 3 53 10 2 3 7 . . . g cm cm 2 g 3 e j . The pellet consists of equal numbers of 2 H and 3 H atoms, so the average molar mass is 2.50 and the total number of atoms is N = × F H G I K J × = × 3 53 10 2 50 6 02 10 8 51 10 7 23 16 . . . . g g mol atoms mol atoms e j . When the pellet is vaporized, the plasma will consist of 2 N particles ( N nuclei and N electrons). The total energy delivered to the plasma is 1.00% of 200 kJ or 2.00 kJ. The temperature of the plasma is found from E N k T = 2 3 2 a f e j B as T E Nk = = × × × = × 3 2 00 10 8 51 10 1 38 10 5 68 10 3 16 23 8 B J 3 J K K . . . . e je j . (b) Each fusion event uses 2 nuclei, so N 2 events will occur. The energy released will be E N Q = F H G I K J = × F H G I K J × = × = 2 8 51 10 2 17 59 1 60 10 1 20 10 120 16 13 5 . . . . MeV J MeV J kJ a f e j . P45.59 (a) The solar-core temperature of 15 MK gives particles enough kinetic energy to overcome the Coulomb-repulsion barrier to 1 1 2 3 H He He e 2 4 + + + + ν , estimated as k e e r e a fa f 2 . The Coulomb barrier to Bethe’s fifth and eight reactions is like k e e r e a fa f 7 , larger by 7 2 times, so the required temperature can be estimated as 7 2 15 10 5 10 6 7 × × K K e j . (b) For 12 1 13 C H N + + Q , Q 1 12 000 000 1 007 825 13 005 739
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern