Chapter 31 - Nuclear Physics and Radioactivity - Problems - Page 900: 18

So Energy required to break all the nuclei in $3g$ of Copper penny is $E=1.55921\times10^{25}MeV$

Atomic mass of copper is given as $62.939598u$ $1u$ is equal to 1.6605\times10^{-27}kg so $62.939598u$ is equal to $62.939598\times 1.6605\times10^{-27}kg=104.51120\times10^{-27}kg$ So mass of $1$ atom of copper is $=104.51120\times10^{-27}kg=1.045112\times10^{-25}kg$ Given mass of copper penny is $m=3.0g$=$3.0\times10^{-3}kg$ now from above calculations $ 1.045112\times10^{-25}kg$ is equal to $1$ Copper atom so $3.0\times10^{-3}kg$ is equal to $\frac{3.0\times10^{-3}}{1.045112\times10^{-25}}$ =$2.87051\times^{22}$atoms So our $3.0g$ copper penny contains $2.87051\times^{22}$ atoms or nucleus Atomic mass of copper is given as $62.939598u$ $^{63}_{29}Cu$ Comparing with $ ^{A}_{Z}X$ we get $A=63$ $Z=29$ number of neutrons $N=A-Z$ $N=63-29=34$ no of neutron in Copper nucleus is $N=34$ we can make Copper by adding $34$ neutrons to $29$ hydrogens so sum of their individual masses is $M_{sum}=34\times m_{n} +29\times m_{H}$ $m_{n}=1.0087u$ $m_{H}=1.007825u$ $M_{sum}=34\times 1.0087u +29\times 1.007825u$ $M_{sum}=63.522725u$ mass defect $\Delta m=M_{sum}$$-Atomic$ $mass$ mass defect $\Delta m=63.522725u-62.939598u$ mass defect $\Delta m=0.583127u$ since $1u$ is equivalent to $931.5MeV$ the binding energy will be $B.E.=0.583127u\times\frac{931.5MeV}{1u}$ binding energy of Copper nucleus will be $B.E.=543.1828MeV$ So if we want to break $1$ copper nucleus to its constituent protons and neutrons we need $543.1828MeV$ energy so to break all the $3g$=$2.87051\times^{22}$ nuclei we need total energy $E= 543.1828MeV\times2.87051\times^{22}$ $E=1559.21\times10^{22}MeV$ $E=1.55921\times10^{25}MeV$