24.129 CP - Chemistry: Molecular Nature Of Matter And Change

Question

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

Step-by-Step Solution

Answer

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

Step 1: Definition of Nuclear binding

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.

Step 2: Find the nuclear binding energy?

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

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

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

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

Step 3: Usinge the formula of atomic mass unit

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

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

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