p43_022 - . 008665 u 1 . 000000 u)(931 . 5 MeV / u) = +8 ....

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22. (a) Table 43-1 gives the atomic mass of 1 Has m =1 . 007825 u. Therefore, the mass excess for 1 His ∆=(1 . 007825 u 1 . 000000 u)(931 . 5MeV / u) = +7 . 29 MeV . (b) The mass of the neutron is given in Sample Problem 43-3. Thus, for the neutron, ∆ = (1
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Unformatted text preview: . 008665 u 1 . 000000 u)(931 . 5 MeV / u) = +8 . 07 MeV . (c) Appealing again to Table 43-1, we obtain, for 120 Sn, = (119 . 902199 u 120 . 000000 u)(931 . 5 MeV / u) = 91 . 10 MeV ....
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