Numerical
Calculate the energy released in the following reaction, given the masses to be
\[\ce{_88^223Ra}\] : 223.0185 u, \[\ce{_82^209Pb}\] : 208.9811 u, \[\ce{_6^14C}\] : 14.00324 u, \[\ce{_92^236U}\] : 236.0456 u, \[\ce{_56^140Ba}\] : 139.9106 u, \[\ce{_36^94Kr}\] : 93.9341 u, \[\ce{_6^11C}\] : 11.01143 u, \[\ce{_5^11B}\] : 11.0093 u. Ignore neutrino energy.
\[\ce{_6^11C -> _5^11B + e^+ + neutrino}\]
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Solution
\[\ce{_6^11C -> _5^11B + e^+ + neutrino}\]
The energy released in this reaction = (ΔM)c2
= [11.01143 - (11.0093 + 0.00055)](931.5} MeV
=1.47177 MeV
Concept: Nuclear Energy
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