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Question
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, \[\ce{_92^36Kr}\] : 93.9341 u, \[\ce{_11^6C}\] : 11.01143 u, \[\ce{_11^5B}\] : 11.0093 u. Ignore neutrino energy.
\[\ce{_88^223Ra -> _82^209Pb + _6^14C}\]
Numerical
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Solution
\[\ce{_88^223Ra -> _82^209Pb + _6^14C}\]
The energy released in this reaction = (ΔM)c2
= [223.0185 - (208.9811+14.00324)](931.5) MeV
= 31.820004 Me V
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