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Calculate the energy released, in MeV, in the reaction " "_(3)^(6)Li + " "_(0)^(1)n to " "_(2)^(4)He + " "_(1)^(3)H Given that mass (" "_(3)^(6)Li) = 6.015126 u, mass (n) = 1.008665 u, m(" "_(2)^(4)He) = 4.002604 and m(" "_(1)^(3)H) = 3.010000. Take 1 u = 931 MeV/c^(2). |
| Answer» SOLUTION :4.782MeV | |