Calculate the binding energy per nucleon for `._(10)^(20)Ne,._(26)^(56)Fe` and `._(98)^(238)U`. Given that mass of neutron is `1.008665 am u`, mass of proton is `1.007825 am u`, mass of `._(10)^(20)Ne` is `19.9924 am u`, mass of `._(26)^(56)Fe` is `55.93492 am u` and mass of `._(92)^(238)U` is `238.050783 am u`.

Answer

`7.57 MeV`

Explanation

Binding energy of nucleus is given by the equation
`B(._Z^AX) =[(A -Z)m_n +Z m_(p) - M(._Z^AX)]c^(2)`
On dividing binding energy by the mass number, we obtain the binding energy per nucleon.
`B(._(10)^(20)Ne) =[10 m_(n) +10 m_(p) - M(._(10)^(20) Ne)]c^(2)`
`=[10 xx 1.008665 +10 xx 1.007825 -55.93492]`
`xx 931.5`
`=492 MeV`
Hence, binding energy per nucleon `=(B(._(26)^(56)Fe))/(56 )=8.79`
`MeV//nuclon`
Binding energy for `._(92)^(238)U` is
`[146m_(n) +92m_(p) -M(._(92)^(238)U)]c^(2)`
`=[146 xx 1.008665 +92 xx 1.007825 -238.050783]`
`xx9315`
`=1802 MeV`
Binding enregy per nucleon
`=(B(._(92)^(238)Fe))/(238)=(1802)/(238) =7.57 MeV`.