Q. Using the following data Mass hydrogen atom = 1.00783 u Mass of neutron = 1.00867 u Mass of nitrogen atom ($\ _{7}N^{14}$)= 14.00307 u The calculated value of the binding energy of the nucleus of the nitrogen atom ($\ _{7}N^{14}$) is close to
A) 56 MeV
B) 98 MeV
C) 104 MeV
D) 112 MeV
Explanation
(c) The binding energy of nucleus may be defined as the energy equivalent to the mass defect of the nucleus.
If $\mathrm{\Delta}m$ is mass defect than according to Einstein’s mass energy relation.
Binding Energy
$= \mathrm{\Delta}\ mc^{2}\ = \ \left\lbrack \left\{ Zm_{p}\ + \ (A - Z)m_{n} \right\rbrack - M \right\rbrack c^{2}$
$= (7\ \times \ 1.00783\ + \ 7\ \times \ 1.00867 - 14.00307)c^{2}$
$or\ \ BE = 0.1124\ \times \ 931.5\ MeV$
or BE = 104.6
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.