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                                    How much energy is released in the following reaction?7Li+p→α+αAtomic mass of 7Li - 7.0160 u and that of 4He - 4.0026 u. | 
                            
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Answer»  Li7 + p → l + α + E ; Li7 = 7.016u α= 4He = 4.0026u ; p = 1.007276 u E = Li7 + P – 2α = (7.016 + 1.007276)u – (2x 4.0026)u = 0.018076 u. =>0.018076 x 931 = 16.828 = 16.83 MeV.  | 
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