What is the nuclear binding energy of a lithium-7nucleus in units of kJ/mol and eV/nucleus? (Mass of a lithium-7atom =7.016003amu.)

Nuclear binding energy is the amount of energy needed to divide an atom's nucleus into its constituent parts. Neutrons and protons, collectively known as nucleons, are the component parts.

4m is the difference in mass between the reactants and the products:

$$\begin{gathered} 4m=m_{products}-m_{reac\tan ts} \\ \text{Masso}fH=3x1.007825=3.0235amu \\ \text{Massof}n=4x1.008665=4.0347amu \\ {\text{Mass}}_{total}=7.0582amu4m=7.0582amu-7.016003amu \\ =0.0421\frac{amu}{mo{l}^{7}Li} \end{gathered}$$

Binding energies are often measured in millions of electron volts, or mega-electron volts (MeV):

$$\begin{gathered} MeV={10}^{6}eV \\ =1.602x{10}^{-13}J \end{gathered}$$

The atomic mass unit is converted to its energy equivalent in electron volts using the following formula:

$$\begin{gathered} \text{1amu}=931.5x{10}^{6}eV=931.5MeV \\ \text{Energy}=\frac{0.0421\frac{amu}{mol{l}^{7}L}}{1atom}\left(\frac{931.5MeV}{1amu}\right)\left(\frac{{10}^{6}eV}{1MeV}\right) \\ \text{Energy}=39216150\frac{eV}{atom} \\ \text{Energy}=\frac{0.0421\frac{amu}{mol{l}^{7}L}}{1atom}\left(\frac{1kg}{1000g}\right){\left(2.99792x{10}^8\frac{m}{s}\right)}^2\left(\frac{1Jg\times k^2}{s^2}\left(\frac{1kJ}{1000J}\right)\right. \\ \text{Energy}=373747741\frac{kJ}{mol} \end{gathered}$$

Therefore, the nuclear binding energy$\text{Energy}=373747741\frac{kJ}{mol}$