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Calculate the disintegration energy Q for fission of `._42Mo^(98)` into two equal fragments `._21Sc^(49)` by bombarding with a neutron. Given that `m(._42Mo^(98))=97.90541u, m(._21Sc^(49))=48.95002u, m_n=1.00867u` |
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Answer» The disintegration energy in fission of `._42Mo^(82)` is given by `Q=(Deltam)xx931MeV=[m(._42Mo^(82))+m_n-2m(._21Sc^(49))]xx931MeV` `=[97.90541+1.00867-2xx48.95002]xx931MeV` `Q=1.01404xx931=944.1MeV` |
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