Question

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. $56MeV$
• 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
$=\Delta \, mc^{2} \, = \, \left[\left\{Z m_{p} \, + \, \left(A - Z\right) m_{n}\right\} - M\right]c^{2}$
$=\left(7 \, \times \, 1.00783 \, + \, 7 \, \times \, 1.00867 - 14.00307\right)c^{2}$
$orBE=0.1124 \, \times \, 931.5 \, MeV$
or $BE=104.6MeV$