What would be the energy required to dissociate completely `1 g` of `Ca-40` into its constituent, particles? Given: Mass of proton `=1.00866 am u`,
Mass of neutron `=1.00866 am u`,
Mass of `Ca-40 =39.97454 am u`, (Take `1 am u =931 MeV`).

Options

A. ` 4.813 xx 10^(24)MeV`
B. ` 4.813 xx 10^(24)eV`
C. ` 4.813 xx 10^(23)MeV`
D. none of these

Answer

a

Explanation

Mass defect,
`Deltam=20(1.007277 +1.00866)-39.97545`
`=40.31874-39.97545`
`=0.34329 a.m.u.`
`:.` Binding energy `=0.34329 xx 931 =319 .6 MeV`
When one atom of `Ca-40` completely dissociates, the energy to be supplied `=319.6 MeV`
`1 g` of `Ca-40` contains `(6.023 xx10^(23))/(40)=1.506 xx 10^(22)` atoms
The energy required for the dissocoation of `1 g` of `Ca-40`
`319.6 xx 1.506 xx10^(22)`
`4.813 xx 10^(24)MeV`.