238 U → 95 Sr+ 140 Xe + 3n 32.26 given the atomic masses to be m(238 U)= 238.050784 u, m(5 Sr)= 94.919388 u, m(140 Xe)= 139.921610 u, and m(n)= 1.008665 u.

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Chapter29: Nuclear Physics
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Calculate the energy released in the following spontaneous fission reaction:

238 U → 95 Sr+
140 Xe + 3n
32.26
given the atomic masses to be m(238 U)= 238.050784 u, m(5 Sr)= 94.919388 u, m(140 Xe)= 139.921610 u, and
m(n)= 1.008665 u.
Transcribed Image Text:238 U → 95 Sr+ 140 Xe + 3n 32.26 given the atomic masses to be m(238 U)= 238.050784 u, m(5 Sr)= 94.919388 u, m(140 Xe)= 139.921610 u, and m(n)= 1.008665 u.
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