Question

Using the following data
Mass of 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. $112 \, MeV$

Answer 1

Answer: (C)

The binding energy of nucleus may be defined as the energy equivalent to the mass defect of the nucleus.
If $\Delta m$ is mass defect than according to Einstein's mass energy relation.
Binding Energy
$=\Delta m c^{2}=Z m_{p}+A-Z m_{n}-M c^{2} $
$= 7 \times 1.00783+7 \times 1.00867-14.00307 c^{2} $
or $ BE =0.1124 \times 931.5\, MeV $
or $ B E \approx 104.6\, MeV$