Two nuclei of mass number 3 combine with anot | vedclass

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(C) The nuclear reaction is given by: $2 	imes X(A=3) + Y(A=4) 	o Z(A=10)$.Total binding energy of reactants $= (2 	imes 3 	imes 5.6 	ext{ MeV})

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$2.2$

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(C) The nuclear reaction is given by: $2 	imes X(A=3) + Y(A=4) 	o Z(A=10)$.Total binding energy of reactants $= (2 	imes 3 	imes 5.6 	ext{ MeV}) + (4 	imes 7.4 	ext{ MeV}) = 33.6 	ext{ MeV} + 29.6 	ext{ MeV} = 63.2 	ext{ MeV}$.Total binding energy of the product $= 10 	imes 6.1 	ext{ MeV} = 61.0 	ext{ MeV}$.The energy change in the process is $\Delta E = E_{	ext{product}} - E_{	ext{reactants}} = 61.0 	ext{ MeV} - 63.2 	ext{ MeV} = -2.2 	ext{ MeV}$.The magnitude of the energy change involved in the process is $2.2 	ext{ MeV}$.Therefore, $\Delta Mc^2 = 2.2 	ext{ MeV}$.

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