Obtain the binding energy of the nuclei `._26Fe^(56)` and `._83Bi^(209)` in units of MeV from the following data: `m(._26Fe^(56))=55.934939a.m.u.` , `m=(._83Bi^(209))=208.980388 a m u`. Which nucleus has greater binding energy per nucleon? Take `1a.m.u =931.5MeV`

Explanation

Atomic mass of `""_(26)^(56)Fe, m_(1)=55.934939" "u`
`""_(26)^(56)Fe` nucleus has 26 protons and (56 − 26) = 30 neutrons
Hence, the mass defect of the nucleus, `trianglem=26xxm_(H)+30xxm_(n)-m_(1)`
where,
Mass of a proton, `m_(H)=1.007825" "u`
Mass of a neutron, `m_(n)=1.008665" "u`
`therefore triangle m=26xx1.007825+30xx1.008665-55.934939`
`=26.20345+30.25995-55.934939`
`=0.528461" "u`
`But" "1u=931.5 MeV//C^(2)`
The binding energy of this nucleus is given as:
`E_(b1)=triangle mc^(2)`
Where,
c = Speed of light
`therefore E_(b1)=0.528461xx931.5((MeV)/(c^(2)))xxc^(2)`
`=492.26 MeV`
Average binding energy per nucleon `=492.26/56=8.79 Mev`
Atomic mass of `_(83)^(209)Bi,m_(2)=208.980388" "u`
`_(83)^(209)Bi` nucleus has 83 protons and `(209 − 83)` 126 neutrons.
Hence, the mass defect of this nucleus is given as:
`triangle m=83xxm_(H)+126xxm_(n)-m_(2)`
Where,
Mass of a proton, `m_(H)=1.007825" "u`
Mass of a neutron, `m_(n)=1.008665" "u`
`trianglem=83xx1.007825+126xx1.008665-208.980388`
`=83.649475+127.091790-208.980388`
`=1.760877" "u`
`But 1u=931.5MeV//c^(2)`
`therefore triangle m=1.760877xx931.5 MeV//c^(2)`
Hence, the binding energy of this nucleus is given as:
`E_(b2)=triangle mc^(2)`
`=1.760877xx931.5((MeV)/(c^(2)))xxc^(2)`
`=1640.26Mev`
Average bindingenergy per nucleon `=(1640.26)/(209)=7.848Mev`