Question

Calculate the binding energy and binding energy per nucleon (in MeV) of a nitrogen nucleus ${(}_7^{14}N)$ from the following data :
Mass of proton = 1.00783 u
Mass of neutron = 1.00867 u
Mass of nitrogen nucleus = 14.00307 u

Answer 1

Nuclear reaction : $7p+7n{\to }_7^{14}N$

Mass defect $\Delta M=7m_p+7m_n-m_{nucleus}$

$\therefore$ $\Delta M=7\left(1.00783\right)+7\left(1.00867\right)-14.00307$ amu

$⟹$ $\Delta M=0.11243$ amu

Thus binding energy $E=\Delta M\times 931.5$ MeV

$\therefore$ $E=0.11243\times 931.5=104.73$ MeV

Binding energy per nucleon $=\frac{104.73}{14}=7.48$ MeV