Question

Question: Calculate the total binding energy in both kilojoules per mole and MeV per atom, and the binding energy per nucleon of the following nuclides using data from table 19.1

1. ${}_{\mathbf{4}}{}^{\mathbf{0}}\mathbf{Be}$
2. ${}_{\mathbf{17}}{}^{\mathbf{35}}\mathbf{Cl}$
   
3. ${}_{\mathbf{22}}{}^{\mathbf{49}}\mathbf{Ti}$
   

Short answer

The binding energy of the required nucleids are given below.

	Binding energy per mole in kilojoules	Binding energy per atom in MeV	Binding energy per nucleon in MeV
${}_4{}^{10}\mathrm{B}$	$-5.8\times {10}^9\mathrm{KJ}/\mathrm{mol}$	$-61.479\mathrm{MeV}$	$-6.14\mathrm{MeV}$
${}_{17}{}^{35}\mathrm{Cl}$	$-2.7\times {10}^{10}\mathrm{KJ}/\mathrm{mol}$	$-282.2\mathrm{MeV}$	$-8.06\mathrm{MeV}$
${}_{22}{}^{49}\mathrm{Ti}$	$-3.8\times {10}^{10}\mathrm{KJ}/\mathrm{mol}$	$-400.54\mathrm{MeV}$	$-8.17\mathrm{MeV}$

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.

$∆\mathrm{E}=∆{\mathrm{mc}}^2$

Where, localid="1660194393977" $∆\mathrm{m}=$mass difference or the mass that has been converted to energy in a nuclear reaction

localid="1660199203459" $∆\mathrm{E}=$energy

c = speed of light

We know,

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

We will use the above relations in further calculations.

Binding energy for Be410

Firstly we have to calculate the mass difference.

$$\begin{gathered} ∆\mathrm{m}={\mathrm{m}}_{\mathrm{atom}-}\left({\mathrm{m}}_{\mathrm{protons}}+{\mathrm{m}}_{\mathrm{neutrons}}\right) \\ =10.01\mathrm{u}-\left(4\times 1.007+6\times 1.008\right)\mathrm{u} \\ =10.01\mathrm{u}-10.076\mathrm{u} \\ =-0.066\mathrm{u} \end{gathered}$$

From Einstein’s mass-energy relation,

$$\begin{gathered} {\text{∆E=∆mc}}^2 \\ ∆\mathrm{E}=-0.66\mathrm{u}\times 1.66\times {10}^{-27}\mathrm{kg}/\mathrm{u}\times {\left(2.99\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2 \\ =-0.98\times {10}^{-11}\mathrm{J} \end{gathered}$$

Binding energy per mole $=-9.8\times {10}^{11}\mathrm{J}\times 6.023\times {10}^{23}=-5.8\times {10}^9\mathrm{KJ}/\mathrm{mol}$

Binding energy per atom $=-0.066\mathrm{u}\times 931.5\mathrm{MeV}=-61.479\mathrm{MeV}$

Binding energy per nucleon $=\frac{-61.479}{10}=-6.147\mathrm{MeV}$

Binding energy for Cl1735

Firstly we have to calculate the mass difference.

$$\begin{gathered} ∆\mathrm{m}={\mathrm{m}}_{\mathrm{atom}-}\left({\mathrm{m}}_{\mathrm{protons}}+{\mathrm{m}}_{\mathrm{neutrons}}\right) \\ =34.96\mathrm{u}-\left(17\times 1.007+18\times 1.008\right)\mathrm{u} \\ =34.96\mathrm{u}-35.263\mathrm{u} \\ =-0.303\mathrm{u} \end{gathered}$$

From Einstein’s mass-energy relation,

localid="1660199741447" $$\begin{gathered} {\text{∆E=∆mc}}^2 \\ ∆\mathrm{E}=-0.303\mathrm{u}\times 1.66\times {10}^{-27}\mathrm{kg}/\mathrm{u}\times {\left(2.99\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2 \\ =-4.49\times {10}^{-11}\mathrm{J} \end{gathered}$$

Binding energy per mole $=-4.49\times {10}^{11}\mathrm{J}\times 6.023\times {10}^{23}=-2.7\times {10}^{10}\mathrm{KJ}/\mathrm{mol}$

Binding energy per atom$=-0.303\mathrm{u}\times 931.5\mathrm{MeV}=-282.82\mathrm{MeV}$

Binding energy per nucleon localid="1660199971879" $=\frac{-282.82}{35}=-8.06\mathrm{MeV}$

Binding energy of Ti2249

Firstly we have to calculate the mass difference.

$$\begin{gathered} ∆\mathrm{m}={\mathrm{m}}_{\mathrm{atom}-}\left({\mathrm{m}}_{\mathrm{protons}}+{\mathrm{m}}_{\mathrm{neutrons}}\right) \\ =48.94\mathrm{u}-\left(22\times 1.007+27\times 1.008\right)\mathrm{u} \\ =48.94\mathrm{u}-49.37\mathrm{u} \\ =-0.43\mathrm{u} \end{gathered}$$

From Einstein’s mass-energy relation,

$$\begin{gathered} {\text{∆E=∆mc}}^2 \\ ∆\mathrm{E}=-0.43\mathrm{u}\times 1.66\times {10}^{-27}\mathrm{kg}/\mathrm{u}\times {\left(2.99\times {10}^8\mathrm{m}/\mathrm{s}\right)}^2 \\ =-6.381\times {10}^{-11}\mathrm{J} \end{gathered}$$

Binding energy per mole$=-6.381\times {10}^{11}\mathrm{J}\times 6.023\times {10}^{23}=-3.8\times {10}^{10}\mathrm{KJ}/\mathrm{mol}$

Binding energy per atom $=-0.43\mathrm{u}\times 931.5\mathrm{MeV}=-400.54\mathrm{MeV}$

Binding energy per nucleon $=\frac{-400.54}{49}=-8.17\mathrm{MeV}$