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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. 431 with Appendix F and/or
G) the copper63 nucleus has 29 protons and 34 neutrons. We use the more accurate values given in
Sample Problem 433:
∆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. 4320:
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 . ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Energy

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