1260_Physics ProblemsTechnical Physics

1260_Physics ProblemsTechnical Physics - Chapter 45 P45.18...

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

View Full Document Right Arrow Icon
Chapter 45 601 P45.18 (a) V F H G I K J 317 10 1609 132 10 6 3 18 mi m 1 mi m 33 ej . mV m M M m mm water H H HO Deuterium H kg m m kg kg kg kg kg 2 2 2 2 2 == × = × = F H G I K J = F H G I K J ×= × × × = × ρ 10 1 32 10 1 32 10 2016 18 015 1 32 10 1 48 10 0 030 0% 0 030 0 10 1 48 10 4 43 10 31 8 2 1 21 20 22 0 1 6 e j bg e j .. . . The number of deuterium nuclei in this mass is N m m × × Deuterium Deuteron kg 2.014 u kg u 443 10 166 10 133 10 16 27 43 . . . af . Since two deuterium nuclei are used per fusion, 1 2 1 2 HH H e 2 4 +→ + Q , the number of events is N 2 663 10 42 The energy released per event is QM M M c =+ = = 2 HHH e u 931.5 MeV u MeV 24 2 2 2 014102 4 002 603 23 8 . b g . The total energy available is then E N Q = F H G I K J × F H G I K J 2 238 160 10 253 10 42 13 31 . . a f MeV J 1 MeV J. (b) The time this energy could possibly meet world requirements is t E × × × F H G I K J P 361 10 1 114 10 1 31 16 9 . ~ J 100 7.00 10 J s s yr 3.16 10 s yr billion years 12 7 . P45.19 (a) Including both ions and electrons, the number of particles in the plasma is
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/15/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

Ask a homework question - tutors are online