Question

Calculate the total binding energy, in both kJ per mole and MeV per atom, and the binding energy per nucleon of the following nuclides using the data from Table 19.1.

(a) ${}_{\mathbf{20}}{}^{\mathbf{40}}\mathbf{Ca}$

(b) ${}_{\mathbf{37}}{}^{\mathbf{87}}\mathbf{Rb}$

(c) ${}_{\mathbf{92}}{}^{\mathbf{238}}\mathbf{U}$

Short answer

Element	Binding energy in kJ	Binding energy in MeV	Binding energy per nucleon
${}_{\text{20}}{}^{\text{40}}\text{Ca}$	$-3.04\times {10}^{-10}KJ/mol$	-316.71 MeV	-7.91 MeV
${}_{\text{37}}{}^{\text{87}}\text{Rb}$	$-6.7\times {10}^{-10}KJ/mol$	-707.0 MeV	-8.12 MeV
${}_{\text{92}}{}^{\text{238}}\text{U}$	$-1.57\times {10}^{-11}KJ/mol$	1639.4 MeV	-5.5 MeV

Step-by-step solution

Mass difference and Binding energy

The mass difference of a nucleus is calculated by subtracting the mass of the protons and neutrons from the total mass of the atom. The Einstein’s Mass- Energy relationship gives an expression to transform the mass into energy. The Mass- Energy equation is given below.

$\text{E} \text{=} \Delta {\text{mc}}^{\text{2}}$

Where, $∆m=$ mass difference or the mass that has been converted to energy in a nuclear reaction

$\text{E =}$ energy

$\text{c =}$ speed of light

We know,

$1 \text{atomic} \text{mass} \text{unit} =931.5 \text{MeV}$

We will use the above relation in further calculations.

Binding energy of  Ca2040

We will first calculate the mass difference of${}_{\text{20}}{}^{\text{40}}\text{Ca}$.

$$\begin{gathered} \Delta m=m_{atom}-\left(m_{protons}+m_{neutrons}\right) \\ =39.96 u-\left(20\times 1.007+20\times 1.008\right) u \\ =\left(39.96-40.3\right) u \\ =-0.34 u \end{gathered}$$

From the mass difference, we will calculate the binding energy using Einstein’s mass-energy relation.

$$\begin{gathered} \Delta E=-0.34 u\times 1.66\times {10}^{-27} kg/u\times \left(2.99\times {10}^{8}m/s^2\right) \\ =-5.06\times {10}^{-11} J \end{gathered}$$

The binding energy for 1 mole ${}_{\text{20}}{}^{\text{40}}\text{Ca}$$=-5.05\times {10}^{-11} J\times 6.023\times {10}^{23}= -3.04\times {10}^{10} KJ/mol$

The energy equivalent to -0.34 u $=-0.34\times 931.5 MeV=-316.71 MeV$

The binding energy of ${}_{\text{20}}{}^{\text{40}}\text{Ca}$ is $-5.06\times {10}^{-11} \text{J}$ or $-316.71 MeV$.

The binding energy per nucleon $=\frac{-316.71}{40}=-7.91 MeV$

Binding energy of  Rb3787

$$\begin{gathered} \Delta m=m_{atom}-\left(m_{protons}+m_{neutrons}\right) \\ =86.90 u-\left(37\times 1.007+50\times 1.008\right) u \\ =\left(86.90-87.659\right) u \\ =-0.759 u \end{gathered}$$

From the mass difference, we will calculate the binding energy using Einstein’s mass-energy relation.

$$\begin{gathered} \Delta E=-0.759 u\times 1.66\times {10}^{-27} kg/u\times \left(2.99\times {10}^{8}m/s^2\right) \\ =-11.26\times {10}^{-11} J \end{gathered}$$

The binding energy for 1 mole ${}_{37}{}^{87}\text{Rb}$ $= -11.26\times {10}^{-11} J\times 6.023\times {10}^{23}= 6.7\times {10}^{10} KJ/mol$

The energy equivalent to $-0.759 u=-0.759\times 931.5 eV=-707.0 MeV$

The binding energy of a ${}_{37}{}^{87}\text{Rb}$nucleus is $-1.126\times {10}^{-10} J or -707.0 MeV$

The binding energy per nucleon $=\frac{-707.0}{87}=-8.12 MeV$

Binding energy of  U92238

$$\begin{gathered} \Delta m=m_{atom}-\left(m_{protons}+m_{neutrons}\right) \\ =238.05 u-\left(92\times 1.007+146\times 1.008\right) u \\ =\left(238.05-239.81\right) u \\ =-1.76 u \end{gathered}$$

From the mass difference, we will calculate the binding energy using Einstein’s mass-energy relation.

$$\begin{gathered} \Delta E=-1.76 u\times 1.66\times {10}^{-27} kg/u\times \left(2.99\times {10}^{8}m/s^2\right) \\ =-26.11\times {10}^{-11} J \end{gathered}$$

The binding energy for 1 mole ${}_{37}{}^{87}\text{Rb}$$= -26.11\times {10}^{-11} J\times 6.023\times {10}^{23}= 1.57\times {10}^{11} KJ/mol$

The energy equivalent to $-1.76 u=-1.76\times 931.5 eV=-1639.4 MeV$

The binding energy of a ${}_{92}{}^{238}\text{U}$ atom is $-2.61\times {10}^{-10} J or -1639.4 MeV$

The binding energy per nucleon $=\frac{-1639.4}{298}=-5.5 MeV$