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Bombardment of lithium with protons gives rise to the following reaction : `._(3)^(7)Li + ._(1)^(1)H rarr 2[._(2)^(4)He]+Q`. Find the Q - value of the reaction. The atomi mases of lithium, proton and helium are 7.016u, 1.0084 and 4.004 u respectively. |
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Answer» Mass of Lithrium = 7.016 u `m_(P)=1.008 u` Mass of Helium = 4.004 u, `u=931.5 MeV` Q = [Total mass of the reactants - Total mass of the products `c^(2)` = [Mass of Lithium `+ m_(P) -` (2 `xx` mass of Helium)] `xx 931.5 MeV` `= [7.016+1.008-2(4.004)]xx931.5 MeV` `=[8.024-8.008]xx931.5 MeV` `therefore` Energy `Q=0.016xx931.5` `= 14.904 MeV` |
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