p44_023

# p44_023 - 58 × 10 24 = 1 72 × 10 25(c One neutron...

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

23. (a) The energy yield of the bomb is E =(66 × 10 3 megaton)(2 . 6 × 10 28 MeV / megaton) = 1 . 72 × 10 27 MeV. At 200 MeV per Fssion event, (1 . 72 × 10 27 MeV) / (200 MeV) = 8 . 58 × 10 24 Fssion events take place. Since only 4 . 0% of the 235 U nuclei originally present undergo Fssion, there must have been (8 . 58 × 10 24 ) / (0 . 040) = 2 . 14 × 10 26 nuclei originally present. The mass of 235 U originally present was (2 . 14 × 10 26 )(235 u)(1 . 661 × 10 27 kg / u) = 83 . 7kg. (b) Two fragments are produced in each Fssion event, so the total number of fragments is 2(8
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . 58 × 10 24 ) = 1 . 72 × 10 25 . (c) One neutron produced in a Fssion event is used to trigger the next Fssion event, so the average number of neutrons released to the environment in each event is 1 . 5. The total number released is (8 . 58 × 10 24 )(1 . 5) = 1 . 29 × 10 25 ....
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