Q24.80P - Chemistry: Molecular Nature Of Matter And Change

Question

Lead-$\mathbf{206}$ is the end product of 238U decay. One ${}^{\mathbf{\text{206}}}\mathbf{\text{Pb}}$ atom has a mass of $\mathbf{\text{205.974440}}$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

$$\begin{gathered} \text{a) 7.926 MeV/nucleon} \\ \text{b) 1632.76 MeV/atom} \\ {\text{c) 1.575×10}}^{\text{11}}\text{ kJ/mol} \end{gathered}$$

Step 1: To calculate the binding energy (a)

-  $\text{1 amu} \text{ = } \text{931.5 MeV}$

- The atomic mass of ${}^{\text{206}}\text{Pb is 205.974440 amu}$

- Atomic number of $\text{Pb}$ is $82$ , so it has $82$ protons

- The number of nucleons is  $206$, therefore,

the number of neutrons is ($206$-$82$) $=124$.

- 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

$$\begin{gathered} ∆\text{m = 205.974440amu - }\left({\text{82×m}}_{\text{p}}{\text{ + 124×m}}_{\text{n}}\right) \\ =\text{205.974440amu -} \left(\text{82×1.007825 amu + 124×1.008665 amu}\right) \\ =\text{ - 1.74167amu} \end{gathered}$$

(a)

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

$\text{BE per nucleon = }\frac{{\displaystyle \text{1.74167 amu×}\frac{\text{931.5 MeV}}{\text{1 amu}}}}{{\displaystyle \text{206 nucleons}}}\text{ = 7.926MeV/ nucleon }$

Step 2: To calculate the binding energy (b)

(b)

The binding energy per atom is

$\text{BE per atom = } \text{7.926 MeV/ nucleon×206 nucleons/atom = } \text{1632.76 MeV/ atom}$

Step 3: To calculate the binding energy (c)

(c)

The binding energy per mole in 

$$\begin{gathered} \text{BE per mol in kJ = } {\text{ BE per atom×6.022×10}}^{\text{23}}\text{ atoms /mol} \\ \text{ = } \text{1632.76 MeV/ atom}{\times 6.022\times 10}^{\text{23}} \text{atoms /mol} \\ ={\text{9.83×10}}^{\text{26 }}\text{MeV/mol×}\frac{{\displaystyle {\text{1.602×10}}^{\text{ - 13}}\text{ J}}}{{\displaystyle \text{1 MeV}}} \\ ={\text{1.575×10}}^{\text{14}}\text{ J/mol×}\frac{\text{1 kJ}}{\text{1000 J}} \\ ={\text{1.575×10}}^{\text{14}}\mathrm{kJ}/\mathrm{mol} \end{gathered}$$