Q24.82P - Chemistry: Molecular Nature Of Matter And Change

Question

Iodine-$\mathbf{\text{131}}$ is one of the most important isotopes used in the diagnosis of thyroid cancer. One atom has a mass of $\mathbf{\text{130.906114}}$amu. Calculate the binding energy 

(a) per nucleon in MeV;

 (b) per atom in MeV;

(c) per mole in kJ.

Step-by-Step Solution

Answer

The binding energies are

$$\begin{gathered} \text{a) 8.422 MeV/nucleon} \\ \text{b) 1103.282 MeV/atom} \\ {\text{c) 1.064×10}}^{\text{11}}\text{ kJ/mol} \end{gathered}$$

Step 1: To calculate the binding energy (a)

- $\text{1 amu = 931.5 MeV}$

- The atomic mass of ${}^{\text{131}}\text{I is 130.906114 amu}$

- Atomic number of $\text{I is 53}$, so it has $\text{53}$ protons

- The number of nucleons is $131$, therefore,

the number of neutrons is ($131$-$53$)$=78$

- The mass of neutron is ${\text{m}}_{\text{n}}\text{ = 1.008665 amu}$

- The mass of proton is ${\text{m}}_{\text{p}}\text{ = 1.007825 amu}$

Now, let us calculate the change in mass

$$\begin{gathered} ∆\text{m = } \text{130.906114 amu - }\left({\text{53×m}}_{\text{p}}{\text{ + 78×m}}_{\text{u}}\right) \\ =\text{130.906114 amu - } \left(\text{53×1.007825 amu+ 78×1.008665 amu}\right) \\ =\text{ - 1.184481 amu} \end{gathered}$$

(a)

The binding energy per nucleon, in $\text{MeV}$ is

$\text{BE per nucleon = }\frac{{\displaystyle \text{1.184481 amu×}\frac{\text{931.5 MeV}}{\text{1 amu}}}}{{\displaystyle \text{131 nucleons}}}\text{ = } \text{8.422 MeV/ nucleon}$

Step 2: To calculate the binding energy (b)

(b)

The binding energy per atom is

$\begin{array}{rcl}\text{BE per atom }& =& \text{8.422 MeV/ nucleon×131 nucleons/atom}\\ & =& \text{1103.282 MeV/ atom}\end{array}$

Step 3: To calculate the binding energy (c)

(c)

The binding energy per mole in $\text{kJ}$

$\begin{array}{rcl}\text{BE per mol in kJ}& =& {\text{BE per atom×6.022×10}}^{\text{23}}\text{ atoms /mol}\\ & =& \text{1103.282 MeV/ atom}{\times 6.022\times 10}^{\text{23}} \text{atoms /mol}\\ & =& {\text{6.644×10}}^{\text{26}}\text{ MeV/mol×}\frac{{\displaystyle {\text{1.602×10}}^{\text{ - 13}}\text{ J}}}{{\displaystyle \text{1 MeV}}}\\ & =& {\text{1.064×10}}^{\text{14 }}\text{J/mol×}\frac{\text{1 kJ}}{\text{1000 J}}\\ & =& {\text{1.064×10}}^{\text{11}}\mathrm{kJ}/\mathrm{mol}\end{array}$