represent simple nuclear reactions by nuclear equations of the form NH eO H714 24 817 11" ++

Mass defect and nuclear binding energy

What is mass defect?

Imagine you have a Lego tower made of many blocks. Each block has a certain weight, but when you glue them together the tower feels lighter because some weight is “lost” in the glue. In a nucleus, the total mass of all protons and neutrons together is slightly larger than the mass of the nucleus itself. That missing mass is called the mass defect and it is converted into binding energy that holds the nucleus together. 🔬

Binding energy (Eb)

The mass defect (Δm) is related to the binding energy by Einstein’s famous equation: $$E_b = \Delta m \, c^2$$ where $c$ is the speed of light ($c \approx 3.00 \times 10^8 \text{ m/s}$). The larger the binding energy per nucleon, the more stable the nucleus. 📚

Simple nuclear reactions

Here are a few common reactions written in nuclear notation:

Reaction Description
$^4_2\text{He} + ^1_0\text{n} \rightarrow ^5_2\text{He}$ Helium nucleus captures a neutron.
$^1_0\text{n} + ^{14}_7\text{N} \rightarrow ^{15}_7\text{N} + \gamma$ Neutron capture by nitrogen, emitting a gamma ray.
$^2_1\text{H} + ^{3}_1\text{H} \rightarrow ^{4}_2\text{He} + ^1_0\text{n}$ Deuterium + tritium fusion (used in fusion research).

Calculating mass defect and binding energy

  1. Step 1: Write the masses of all reactants and products (in atomic mass units, u). Example: For $^4_2\text{He}$ (mass = 4.0026 u) + $^1_0\text{n}$ (mass = 1.0087 u) → $^5_2\text{He}$ (mass = 5.0129 u).
  2. Step 2: Calculate the total mass of reactants: $M_{\text{reactants}} = 4.0026 + 1.0087 = 5.0113 \text{ u}$.
  3. Step 3: Calculate the total mass of products: $M_{\text{products}} = 5.0129 \text{ u}$.
  4. Step 4: Find the mass defect: $$\Delta m = M_{\text{reactants}} - M_{\text{products}} = 5.0113 - 5.0129 = -0.0016 \text{ u}$$ (A negative value indicates the reaction releases energy; take the absolute value for binding energy).
  5. Step 5: Convert Δm to kilograms: $1 \text{ u} = 1.6605 \times 10^{-27} \text{ kg}$ → $\Delta m = 0.0016 \times 1.6605 \times 10^{-27} \text{ kg} = 2.66 \times 10^{-30} \text{ kg}$.
  6. Step 6: Calculate binding energy: $$E_b = \Delta m \, c^2 = 2.66 \times 10^{-30} \times (3.00 \times 10^8)^2 \approx 2.39 \times 10^{-13} \text{ J}$$ Convert to MeV: $1 \text{ MeV} = 1.602 \times 10^{-13} \text{ J}$ → $E_b \approx 1.49 \text{ MeV}$ per reaction.

Remember: the larger the binding energy per nucleon, the more stable the nucleus. 💡

Exam tips for A-Level Physics

  • Show all steps clearly – examiners look for the process, not just the final answer.
  • Use the correct units: atomic mass units (u), kilograms (kg), joules (J), or mega‑electronvolts (MeV).
  • Remember the conversion: $1 \text{ u} = 931.5 \text{ MeV}/c^2$ – this shortcut is handy for quick calculations.
  • When comparing binding energies, plot a graph of binding energy per nucleon vs. mass number to identify the most stable nuclei.
  • Use diagrams or tables to organise data – a clear table often earns extra marks.
  • Check that your final answer is in the required format (e.g., “$E_b = 1.49 \text{ MeV}$ per reaction”).

Good luck, and keep practising! 🚀

Revision

Log in to practice.

10 views 0 suggestions