The mass of a {}_{3}^{7}Li nucleus is 0.042 , | vedclass

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(B) For the ${}_{3}^{7}Li$ nucleus,the mass defect is given as $\Delta M = 0.042 \, u$.We know that $1 \, u = 931.5 \, MeV/c^2$.Therefore,the total bi

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$5.6$

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(B) For the ${}_{3}^{7}Li$ nucleus,the mass defect is given as $\Delta M = 0.042 \, u$.We know that $1 \, u = 931.5 \, MeV/c^2$.Therefore,the total binding energy $E_b$ is calculated as:$E_b = \Delta M 	imes 931.5 \, MeV/u = 0.042 	imes 931.5 \, MeV \approx 39.123 \, MeV$.The number of nucleons $A$ in ${}_{3}^{7}Li$ is $7$.The binding energy per nucleon is given by $E_{bn} = \frac{E_b}{A}$.$E_{bn} = \frac{39.123 \, MeV}{7} \approx 5.589 \, MeV \approx 5.6 \, MeV$.

The mass of a ${}_{3}^{7}Li$ nucleus is $0.042 \, u$ less than the sum of the masses of all its nucleons. The binding energy per nucleon of ${}_{3}^{7}Li$ nucleus is nearly...........$MeV$.

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