The isotopic mass of {}_{3}^{7}Li is 7.016005 | vedclass

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(B) The binding energy $(BE)$ is calculated using the mass defect $(\Delta m)$: $\Delta m = [Z 	imes M_{H}   N 	imes M_{n}] - M_{nucleus}$ Here,$Z =

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$39.2$

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(B) The binding energy $(BE)$ is calculated using the mass defect $(\Delta m)$: $\Delta m = [Z 	imes M_{H}   N 	imes M_{n}] - M_{nucleus}$ Here,$Z = 3$ (protons),$N = 4$ (neutrons),$M_{H} = 1.007825 \ u$,$M_{n} = 1.008665 \ u$,and $M_{Li} = 7.016005 \ u$. $\Delta m = [3 	imes 1.007825   4 	imes 1.008665] - 7.016005$ $\Delta m = [3.023475   4.034660] - 7.016005 = 7.058135 - 7.016005 = 0.04213 \ u$ $BE = \Delta m 	imes 931 \ MeV/u = 0.04213 	imes 931 \approx 39.2 \ MeV$.

The isotopic mass of ${}_{3}^{7}Li$ is $7.016005 \ u$ and those of the $H$ atom and neutron are respectively $1.007825 \ u$ and $1.008665 \ u$. Then,the binding energy of the ${}_{3}^{7}Li$ nucleus is: (in $MeV$)

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