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If mass of proton =1.007825 u, mass of n...

If mass of proton =1.007825 u, mass of neutron =1.008665 u, and mass of `._(3)Li^(7)` is 7.01599u, and 1a.m.u. =931MeV, what is BE/nucleon of `._(3)Li^(7)`.

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The correct Answer is:
`5.61MeV//N`

`._(3)Li(7)` has 3 protons and 4 neutrons.
`Deltam=(3xx1.007825+4xx1.008665-7.01599)u`
`=0.042145u`
Total BE=`Deltamxx931.5MeV`
`=39.2580MeV`
BE/nucleon=`39.2580/7=5.61MeV//N`
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