p44_010 - 10. First, we figure out the mass of U-235 in...

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

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

Unformatted text preview: 10. First, we figure out the mass of U-235 in the sample (assuming “3.0%” refers to the proportion by weight as opposed to proportion by number of atoms): MU−235 = = (97%)m238 + (3.0%)m235 (97%)m238 + (3.0%)m235 + 2m16 0.97(238) + 0.030(235) (0.030)(1000 g) 0.97(238) + 0.030(235) + 2(16.0) (3.0%)Msam = 26.4 g . Next, this uses some of the ideas illustrated in Sample Problem 43-5; our notation is similar to that used in that example. The number of 235 U nuclei is N235 = (26.4 g)(6.02 × 1023 /mol) = 6.77 × 1022 . 235 g/mol If all the U-235 nuclei fission, the energy release (using the result of Eq. 44-6) is N235 Qfission = 6.77 × 1022 (200 MeV) = 1.35 × 1025 MeV = 2.17 × 1012 J . Keeping in mind that a Watt is a Joule per second, the time that this much energy can keep a 100-W lamp burning is found to be t= 2.17 × 1012 J = 2.17 × 1010 s ≈ 690 y . 100 W If we had instead used the Q = 208 MeV value from Sample Problem 44-1, then our result would have been 715 y, which perhaps suggests that our result is meaningful to just one significant figure (“roughly 700 years”). ...
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online