p44_034 - 10 23 mol = 1 5 × 10 26 Multiplying this by Q...

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

View Full Document Right Arrow Icon
34. We are given the energy release per fusion ( Q =3 . 27 MeV = 5 . 24 × 10 13 J) and that a pair of deuterium atoms are consumed in each fusion event. To Fnd how many pairs of deuterium atoms are in the sample, we adapt Eq. 43-20: N d pairs = M sam 2 M d N A = µ 1000 g 2(2 . 0g / mol) ( 6 . 02 ×
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 10 23 / mol ) = 1 . 5 × 10 26 . Multiplying this by Q gives the total energy released: 7 . 9 × 10 13 J. Keeping in mind that a Watt is a Joule per second, we have t = 7 . 9 × 10 13 J 100 W = 7 . 9 × 10 11 s = 2 . 5 × 10 4 y ....
View Full Document

Ask a homework question - tutors are online