Question

Using the following data
Mass hydrogen atom = $1.00783u$
Mass of neutron = $1.00867u$
Mass of nitrogen atom $\left({}_7{N}^{14}\right)=14.00307u$
The calculated value of the binding energy of the nucleus of the nitrogen atom $\left({}_7{N}^{14}\right)$ is close to

• A. $56MeV$
• B. $98MeV$
• C. $104MeV$
• D. $112MeV$

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 then according to Einstein's mass-energy relation.
Binding Energy
$=\Delta m{c}^2=\left[\left\{Z{m}_p+\left(A-Z\right)m_n\right\}-M\right]c^2$
$=\left(7\times 1.00783+7\times 1.00867-14.00307\right)c^2$
$orBE=0.1124\times 931.5MeV$
or $BE=104.6MeV$