p43_023

# p43_023 - 23. We ﬁrst “separate” all the nucleons in...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 23. We ﬁrst “separate” all the nucleons in one copper nucleus (which amounts to simply calculating the nuclear binding energy) and then ﬁgure the number of nuclei in the penny (so that we can multiply the two numbers and obtain the result). To begin, we note that (using Eq. 43-1 with Appendix F and/or G) the copper-63 nucleus has 29 protons and 34 neutrons. We use the more accurate values given in Sample Problem 43-3: ∆Ebe = (29(1.007825 u) + 34(1.008665 u) − 62.92960 u) (931.5 MeV/u) = 551.4 MeV . To ﬁgure the number of nuclei (or, equivalently, the number of atoms), we adapt Eq. 43-20: NCu = 3.0 g 62.92960 g/mol 6.02 × 1023 atoms/mol ≈ 2.9 × 1022 atoms . Therefore, the total energy needed is NCu ∆Ebe = (551.4 MeV) 2.9 × 1022 = 1.6 × 1025 MeV . ...
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