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

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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. 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 . ...
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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.

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