Answer
so mass defect $\Delta m= 0.24089u$
Work Step by Step
Given that binding energy of nucleus is $= 225.0MeV=225.0\times10^{6}eV$
or binding energy $=2.25\times10^{8}eV$
1 $eV$ is $1.6\times10^{-19}J$
so Binding energy $=2.25\times10^{8}\times 1.6\times10^{-19}J$
Binding energy $=3.6\times10^{-11}J$
From equation 31.3
Binding energy $=B.E. =$ (mass defect)$\times c^2$=$\Delta mc^2$
mass defect $= \frac{B.E.}{c^2}$
putting the value of Binding energy $=3.6\times10^{-11}J$ and $c=3\times10^{8}m/s$
mass defect $\Delta m= \frac{3.6\times10^{-11}J}{(3\times10^{8}m/s)^2}$
mass defect $\Delta m=0.4\times10^{-27}kg$
$1 u $ is equal to $1.6605\times10^{-27}kg$
$1.6605\times10^{-27}kg$ is equal to $1 u $
$1kg$ is equal to $\frac{1}{1.6605\times10^{-27}}u$
so $0.4\times10^{-27}kg$ will be $\frac{0.4\times10^{-27}}{1.6605\times10^{-27}}u=0.24089u$
so mass defect $\Delta m= 0.24089u$
