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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`).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» Correct Answer - a 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`. |
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