p43_013

# p43_013 - m = (94)(1 . 00783 u) + (239 94)(1 . 00867 u)...

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13. The binding energy is given by ∆ E be =[ Zm H +( A Z ) m n M Pu ] c 2 ,where Z is the atomic number (number of protons), A is the mass number (number of nucleons), m H is the mass of a hydrogen atom, m n is the mass of a neutron, and M Pu is the mass of a 239 94 Pu atom. In principle, nuclear masses should be used, but the mass of the Z electrons included in ZM H is canceled by the mass of the Z electrons included in M Pu , so the result is the same. First, we calculate the mass di±erence in atomic mass units:
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Unformatted text preview: m = (94)(1 . 00783 u) + (239 94)(1 . 00867 u) (239 . 05216 u) = 1 . 94101 u. Since 1 u is equivalent to 931 . 5 MeV, E be = (1 . 94101 u)(931 . 5 MeV / u) = 1808 MeV. Since there are 239 nucleons, the binding energy per nucleon is E ben = E/A = (1808 MeV) / 239 = 7 . 56 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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