Q.29. Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass number 3, 4 and 10 are 5.6 MeV, 7.4

Question

Q.29. Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass number 3, 4 and 10 are 5.6 MeV, 7.4 MeV and 6.1 MeV, respectively, then in the process,  = ________ MeV.

FirstInTest · Accepted Answer

Correct answer: C. Sol. The reaction is $2 	imes (A=3) + (A=4) ightarrow (A=10)$. Total binding energy of reactants = $2 	imes 3 	imes 5.6 + 4 	imes 7.4 = 33.6 + 29.6 = 63.2 	ext{ MeV}$. Total binding energy of product = $10 	imes 6.1 = 61 	ext{ MeV}$. Energy released ($\Delta Mc^2$) = Binding Energy of Reactants - Binding Energy of Products = $63.2 - 61 = 2.2 	ext{ MeV}$.

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