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. `4.831 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

(a) Mass defect
`Delta m = 20(1.007277 + 1.00866) - 39.97545`
`=40.31874 - 39.97545`
`=0.34329` am u
`:.` Binding energy
`=0.34329 xx 931 xx319.6 MeV`
When one atom of `Ca-40` completely dissociates, the energy to be supplied `= 319.6 MeV`
Number of atoms in `1 g` of `Ca -40 = (6.023 xx 10^23)/(40)`
`=1.506 xx 10^22`
The energy required for the dissociates of `1 g` of `Ca-40`
`=319.6 xx 1.506 xx 10^22`
`=4.813 xx 10^24 MeV`.