understand the equivalence between energy and mass as represented by E = mc2 and recall and use this equation

Mass Defect and Nuclear Binding Energy

What is Mass Defect? 🔬

When protons and neutrons (collectively called nucleons) stick together to form a nucleus, the total mass of the nucleus is a little less than the sum of the individual masses.

Think of building a LEGO tower: each brick has a weight, but once the tower is complete, the whole structure feels a bit lighter because some of the “extra weight” is lost in the glue that holds the bricks together. That missing weight is the mass defect (Δm).

Mathematically:

\$\Delta m = Z\,mp + N\,mn - M_{\text{nucleus}}\$

where

\$Z\$ = number of protons,

\$N\$ = number of neutrons,

\$m_p\$ = mass of a proton,

\$m_n\$ = mass of a neutron,

\$M_{\text{nucleus}}\$ = measured mass of the nucleus.

Binding Energy – The Energy That Holds the Nucleus Together ⚛️

The missing mass is not lost forever – it’s converted into energy that keeps the nucleus stable.

This energy is called the binding energy (\$E_B\$) and is given by Einstein’s famous equation:

\$E_B = \Delta m\,c^2\$

where \$c\$ is the speed of light (\$3.00\times10^8\ \text{m/s}\$).

Because nuclear masses are usually expressed in atomic mass units (u), we use the conversion factor:

\$1\ \text{u} = 931.5\ \text{MeV}/c^2\$

so that

\$E_B\ (\text{MeV}) = \Delta m\ (\text{u}) \times 931.5.\$

Example: Helium‑4 (⁴He) 💡

  1. Helium‑4 has 2 protons and 2 neutrons.

  2. Masses (in u):

    • Proton \$m_p = 1.007276\$

    • Neutron \$m_n = 1.008665\$

    • Helium‑4 nucleus \$M_{\text{He}} = 4.002603\$

  3. Calculate the mass of the separate nucleons:

    \$M{\text{separate}} = 2mp + 2m_n = 2(1.007276) + 2(1.008665) = 4.031882\ \text{u}.\$

  4. Mass defect:

    \$\Delta m = M{\text{separate}} - M{\text{He}} = 4.031882 - 4.002603 = 0.029279\ \text{u}.\$

  5. Binding energy:

    \$E_B = 0.029279 \times 931.5 \approx 27.3\ \text{MeV}.\$

ParticleMass (u)Mass (kg)
Proton1.0072761.6726×10⁻²⁷ kg
Neutron1.0086651.6749×10⁻²⁷ kg
⁴He nucleus4.0026036.6465×10⁻²⁷ kg

Using E = mc² in Everyday Units ⚡

If you know the mass defect in kilograms, you can find the energy in joules directly:

\$E = \Delta m\,c^2.\$

For the Helium‑4 example:

\$\Delta m = 0.029279\ \text{u} \times 1.6605\times10^{-27}\ \text{kg/u} = 4.86\times10^{-29}\ \text{kg}.\$

Then

\$E = 4.86\times10^{-29}\ \text{kg} \times (3.00\times10^8\ \text{m/s})^2 \approx 4.4\times10^{-12}\ \text{J}.\$

That’s about the energy released when a tiny amount of matter turns into energy – a huge amount compared to everyday chemical reactions! 🌍

Quick Practice Problem 🎯

A nucleus has 3 protons and 4 neutrons.

Given:

\$mp = 1.007276\ \text{u}\$, \$mn = 1.008665\ \text{u}\$, \$M_{\text{nucleus}} = 7.016929\ \text{u}\$.

Calculate:

  1. Mass defect (Δm).

  2. Binding energy in MeV.

Hints:

1. Use the formula for Δm.

2. Multiply Δm by 931.5 MeV/u to get \$E_B\$.

Key Takeaways 📌

  • The mass of a nucleus is slightly less than the sum of its protons and neutrons.

  • That missing mass is converted into binding energy, keeping the nucleus stable.

  • Einstein’s equation \$E = mc^2\$ links mass and energy – a tiny mass can release a huge amount of energy.

  • Use the conversion 1 u = 931.5 MeV/\$c^2\$ to switch between mass units and energy units.