Question

Calculate the (i) mass defect, (ii) binding energy and (iii) the binding energy per nucleon for a ${}_6{C}^{12}$ nucleus. Nuclear mass of ${}_6{C}^{12}=12.000000$ a.m.u., mass of hydrogen nucleus $=1.007825$ a.m.u. and mass of neutron $=1.008665$ a.m.u.

Answer 1

Given,

Mass of one proton = 1.007825 a.m.u

Mass of one neutron= 1.008665 a.m.u

Nuclear mass of ${}_6{C}^{12}$ = 12 a.m.u

(i) ${}_6{C}^{12}$ has 6 proton, 6 electron and 6 neutron.

Mass of nucleus = Mass of 6 proton + Mass of 6 neutron

$=(6\times 1.007825)+(6\times 1.008665)$

$=12.09894u$

Mass defect $\left(\Delta m\right)$ $=12.09849u-12u=0.098931u$

(ii) Binding Energy = Mass Defect $\times$ $931.5MeV$

$=0.09849u\times 931.5MeV/u=92.15MeV$

(iii) Binding Energy per Nucleon= Binding Energy / Number of Nucleons

Number of nucleon = Mass Number = 12

So, Binding Energy per nucleon $=\frac{92.15}{12}=7.68MeV$