Question

What would be the energy required to dissociate completely $1g$ of $C{a}^{40}$ into its constituent particles?
Given: Mass of proton $=1.007277amu$,
Mass of neutron $=1.00866amu$,
Mass of $Ca-40=39.97545amu$
(Take $1amu=931MeV)$

• A. $4.813\times {10}^{24}MeV$
• B. $4.813\times {10}^{24}eV$
• C. $4.813\times {10}^{23}MeV$
• D. none of the above

Answer 1

The correct option is A $4.813\times {10}^{24}MeV$

$Ca=20p+20n$

For $1$ atom of $Ca$:
Mass Defect $=20\times 1.007277+20\times 1.00866-39.97545=40.31-39.97545=0.34329amu$

Energy released for $1$ atom of $Ca=319.6MeV$
Energy release for $1gm$ of $Ca=\frac{1}{39.975}\times 6.023\times {10}^{23}\times 319.6$$=4.81\times {10}^{24}MeV$