Problem Statement
Determine the binding energy per nucleon of the isotope $^{160}X(1)$ ($Z = 1$, $N = 159$). The atomic mass excess gives the bare nuclear mass as $M = 160.09300\,\text{u}$. Using $m_p = 1.00728\,\text{u}$ and $m_n = 1.00867\,\text{u}$, find (a) the mass defect $\Delta m$, (b) the total binding energy $E_b$, and (c) the binding energy per nucleon $E_b/A$.
Given Information
- $A = 160$ — mass number
- $Z = 1$, $N = 159$ — proton and neutron numbers
- $m_p = 1.00728\,\text{u}$, $m_n = 1.00867\,\text{u}$
- $M(^{160}X(1)) = 160.09300\,\text{u}$
- $1\,\text{u} = 931.5\,\text{MeV}/c^2$
Physical Concepts & Formulas
Nuclear binding energy is the energy that would be required to disassemble a nucleus into free nucleons. By Einstein’s relation, it equals $c^2$ times the mass defect — the difference between the sum of free constituent masses and the actual nuclear mass. Dividing by $A$ gives the binding energy per nucleon, whose curve peaks near $^{56}$Fe and explains why energy is released both by fusion of light nuclei and by fission of heavy ones.
- $\Delta m = (Z m_p + N m_n) – M$ — mass defect
- $E_b = \Delta m \cdot c^2 = \Delta m \cdot 931.5\,\text{MeV}/\text{u}$
- $E_b/A$ — binding energy per nucleon
- $E_b \approx a_V A – a_S A^{2/3} – a_C \dfrac{Z^2}{A^{1/3}} – a_A \dfrac{(N-Z)^2}{A}$ — semi-empirical (Weizsäcker)
Step-by-Step Solution
Step 1 — Sum of free nucleon masses: Add the masses of $Z$ free protons and $N$ free neutrons.
$$\Sigma = 1\cdot1.00728 + 159\cdot1.00867 = 161.38581\,\text{u}$$
Step 2 — Compute mass defect: Subtract the measured nuclear mass.
$$\Delta m = 161.38581 – 160.09300 = 1.29281\,\text{u}$$
Step 3 — Convert to MeV and divide: Apply the $c^2$ conversion factor and divide by $A$ to get the per-nucleon binding.
$$E_b = 1.29281 \cdot 931.5 = 1204.25\,\text{MeV}, \quad \dfrac{E_b}{A} = 7.527\,\text{MeV}$$
Worked Calculation
$$E_b = 1204.2\,\text{MeV}, \quad E_b/A = 7.53\,\text{MeV/nucleon}$$
Answer
$$\boxed{\dfrac{E_b}{A} \approx 7.53\,\text{MeV/nucleon}}$$
This result follows directly from the physical principles laid out above; the numerical magnitude is consistent with the expected scale for this class of binding problem.
Physical Interpretation
For $^{160}X(1)$ the per-nucleon binding is about 7.5 MeV, characteristic of medium-mass nuclei. The empirical curve peaks at $^{56}$Fe near 8.79 MeV/nucleon — this is why nature stops fusing in stellar cores once an iron ash builds up.
Binding energy per nucleon is the single most predictive number in nuclear physics: it controls nuclear stability, the energetics of fission and fusion, the abundance pattern from stellar nucleosynthesis, and the design of every reactor and weapon. The semi-empirical Weizsäcker formula reproduces the curve from five terms — volume, surface, Coulomb, asymmetry, and pairing.
If $A$ were doubled deep into the heavy regime, Coulomb repulsion would dominate and $E_b/A$ would drop below 7.5 MeV, making the nucleus prone to spontaneous fission — precisely what happens beyond $^{238}$U. Conversely, very light nuclei have low $E_b/A$ because too many nucleons sit on the surface.
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