Binding energy of iron

Problem

The nucleus $^{56}\text{Fe}$ has mass $55{,}9349\ \text{u}$. Calculate the mass defect relative to its separate constituents — $26$ protons ($m_p = 1{,}00728\ \text{u}$) and $30$ neutrons ($m_n = 1{,}00867\ \text{u}$) — and hence the binding energy per nucleon. Compare the result with the maximum of the Aston curve.

Solution

Mass defect. This is the difference between the mass of the free nucleons and that of the bound nucleus:

$\Delta m = 26\,m_p + 30\,m_n - m_{\text{nucleus}} = 26\cdot 1{,}00728 + 30\cdot 1{,}00867 - 55{,}9349$

$\Delta m = 26{,}18928 + 30{,}2601 - 55{,}9349 = 0{,}51448\ \text{u}$

Total binding energy. Using the equivalence $1\ \text{u} \leftrightarrow 931{,}5\ \text{MeV}$:

$B = \Delta m\cdot 931{,}5\ \frac{\text{MeV}}{\text{u}} = 0{,}51448\cdot 931{,}5 \approx 479{,}2\ \text{MeV}$

Binding energy per nucleon. Dividing by the mass number $A = 56$:

$\frac{B}{A} = \frac{479{,}2}{56} \approx \ev{\frac{B}{A} \approx 8{,}56\ \text{MeV/nucleon}}$

This value is close to the maximum of the Aston curve ($\sim 8{,}8\ \text{MeV/nucleon}$), fully consistent with the fact that iron is among the most stable nuclei found in nature: it is the “bottom” of the energy valley towards which both the fusion of light nuclei and the fission of heavy ones tend.

Error in the textbook's answer key $\approx 7{,}07\ \text{MeV/nucleon}$, which is incorrect: with the data provided the calculation gives $8{,}56\ \text{MeV/nucleon}$, consistent with the peak of the Aston curve. The correct value $8{,}56\ \text{MeV/nucleon}$ is adopted here.

The textbook’s answer key gives

Collegamenti

Argomenti: Fisica nucleare Concetti: Energia di legame nucleare · Equivalenza massa-energia Oggetti: Nucleo atomico