Answer
binding energy will be $ B.E.=39.376MeV$
Work Step by Step
$^{7}_{3}Li$
Comparing with $ ^{A}_{Z}X$
we get
$A=7$
$Z=3$
number of neutrons $N=A-Z$
$N=7-3=4$
no of neutron in lithium nucleus is $N=4$
we can make lithium by adding four neutrons to three hydrogens
so sum of their individual masses is
$M_{sum}=4\times m_{n} +3\times m_{H}$
$m_{n}=1.0087u$
$m_{H}=1.007825u$
$M_{sum}=4\times 10087u +3\times 1.007825u$
$M_{sum}=7.058275u$
mass defect $\Delta m=M_{sum}$$-Atomic$ $mass$
mass defect $\Delta m=7.058275u-7.016003u$
mass defect $\Delta m=0.042272u$
since $1u$ is equivalent to $931.5MeV$
the binding energy will be $B.E.=0.042272u\times\frac{931.5MeV}{1u}$
binding energy will be $B.E.=39.376MeV$
