Q24.81P - Chemistry: Molecular Nature Of Matter And Change

Question

Cobalt-$\mathbf{\text{59}}$ is the only stable isotope of this transition metal. One ${}^{\mathbf{\text{59}}}\mathbf{\text{Co}}$ atom has a mass of $\mathbf{\text{58.933198}}$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

$$\begin{gathered} \text{a) 8.768 MeV/nucleon} \\ \text{b) 517.312 MeV/atom} \\ {\text{c) 4.99×10}}^{\text{10}}\text{ kJ/mol} \end{gathered}$$

Step 1: To calculate the binding energy (a)

- $\text{1 amu = 931.5 MeV}$

- The atomic mass of ${}^{\text{59}}\text{Co is 58.933198 amu}$

- Atomic number of Co is $27$, so it has $27$ protons

- The number of nucleons is $59$, therefore,

the number of neutrons is ($59$-$27$) $=32$.

- 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

$$\begin{gathered} ∆\text{m = } \text{58.933198 amu - }\left({\text{27×m}}_{\text{p}}{\text{ + 32×m}}_{\text{n}}\right) \\ =\text{58.933198 amu - }\left(\text{27×1.007825 amu + 32×1.008665 amu}\right) \\ =\text{ - 0.555357 amu} \end{gathered}$$

(a)

The binding energy per nucleon, in MeV is

$\text{BE per nucleon = }\frac{{\displaystyle \text{0.555357 amu×}\frac{\text{931.5 MeV}}{\text{1 amu}}}}{{\displaystyle \text{59 nucleons}}}\text{ = } \text{8.768 MeV/ nucleon}$

Step 2: To calculate the binding energy (b)

(b)

The binding energy per atom is

$$\begin{gathered} \text{BE per atom = } \text{8.768 MeV/ nucleon×59 nucleons/atom } \\ \text{ = } \text{517.312 MeV/ atom} \end{gathered}$$

Step 3: To calculate the binding energy (c)

(c)

The binding energy per mole in kJ

$$\begin{gathered} \text{BE per mol in k.J = } {\text{ BE per atom×6.022×10}}^{\text{23}}\text{ atoms /mol} \\ =\text{517.312 MeV/ atom}{\times 6.022\times 10}^{\text{23}} \text{atoms /mol} \\ ={\text{3.115×10}}^{\text{26}}\text{ MeV/mol×}\frac{{\displaystyle {\text{1.602×10}}^{\text{ - 13}}\text{ J}}}{{\displaystyle \text{1 MeV}}} \\ ={\text{4.99×10}}^{\text{13 }}\text{J/mol×}\frac{\text{1 kJ}}{\text{1000 J}} \\ ={\text{4.99×10}}^{\text{10}}\mathrm{kJ}/\mathrm{mol} \end{gathered}$$