Question

Obtain the binding energy of the nuclei ${}_{26}F{e}^{56}$ and ${}_{83}B{i}^{209}$ in units of MeV from the following data: mass of hydrogen atom $=1.007825$ a.m.u., mass of neutron $=1.08665$ a.m.u.; mass of ${}_{26}F{e}^{56}$ atom $=55.934939$ a.m.u. and mass of ${}_{83}B{i}^{209}$ atom $=208.980388$a.m.u. Also calculate binding energy per nucleon in the two cases.

Answer 1

(i) ${}_{26}F{e}^{56}$ nucleus contains 26 protons .

Number of neutrons $=(56-26)=30$ neutrons

Now,

Mass of 26 protons$=26$ X $1.007825=26.20345u$

Mass of 30 neutrons$=30$ X $1.008665=30.25995u$

Total mass of 56 nucleons$=56.46340u$

Mass of ${}_{26}Fe56$ nucleus$=55.934939u$

Therefore,

Mass defect, $△m$ $=56.46340-55.934939=0.528461u$

Total Binding Energy$=0.528461$ X $931.5Mev=492.26MeV$

Average binding energy per nucleon$=\frac{492.26}{56}=8.790MeV$

(ii) ${}_{83}B{i}^{209}$ nucleus contains 83 protons

Number of neutrons$=209-83=126neutrons$

Now,

Mass of 83 protons$=83$ X $1.007825=83.649475amu$

Mass of 126 neutrons$=126X1.008665=127.091790amu$

Therefore, total mass of nucleons$=83.649475+127.091790=210.741260amu$

Given, mass of nucleus$=208.980388amu$

Now, mass defect $△m$ $=210.741260-208.980388=1.760872$

Total binding energy$=1.760872$ X $931.5=1640.26MeV$

Therefore, average binding energy per nucleon = $\frac{1640.26}{209}=7.848Mev$