Mass Defect and Binding Energy

Mass Defect and Binding Energy

One of the most profound results of nuclear physics is that a nucleus weighs less than the sum of its parts. This missing mass is converted into the energy that holds the nucleus together.

Mass Defect

For a nucleus with $Z$ protons and $N$ neutrons, the mass defect is:

$\Delta m = Z \cdot m_p + N \cdot m_n - M_{\text{nucleus}}$

where all masses are in atomic mass units (u):

• $m_p = 1.007276$ u (proton mass)
• $m_n = 1.008665$ u (neutron mass)
• $M_{\text{nucleus}}$ is the measured nuclear mass

Binding Energy

By Einstein's $E = mc^2$, the mass defect corresponds to an energy:

$BE = \Delta m \cdot 931.494 \text{ MeV/u}$

The conversion factor $931.494$ MeV/u comes from $1 \text{ u} \cdot c^2$.

Binding Energy Per Nucleon

The binding energy per nucleon $BE/A$ is a measure of nuclear stability:

$\frac{BE}{A} = \frac{\Delta m \cdot 931.494}{Z + N}$

Iron-56 sits at the peak of the binding energy curve — lighter nuclei release energy by fusion, heavier nuclei by fission.

Your Task

Implement three functions. All constants must be defined inside each function.

• mass_defect(Z, N, M_nucleus_u) — returns $\Delta m$ in u
• binding_energy_MeV(Z, N, M_nucleus_u) — returns $BE$ in MeV
• binding_energy_per_nucleon(Z, N, M_nucleus_u) — returns $BE/A$ in MeV/nucleon

Use $m_p = 1.007276$ u, $m_n = 1.008665$ u, and $1 \text{ u} = 931.494$ MeV/$c^2$.