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Mass Defect, Nuclear Binding Energy & Simple Nuclear Reactions (Cambridge A‑Level Physics 9702 – Topic 23.1)

Learning Objectives

• Explain why a nucleus exists (Rutherford α‑particle scattering) and describe the simple nuclear model (protons + neutrons + electrons).
• Define nucleon number (A) and proton number (Z) and distinguish them from “mass number”.
• Define mass defect and show why it occurs.
• Calculate the binding energy of a nucleus using  Δm c².
• Convert mass defect from atomic mass units to mega‑electron‑volts (MeV) using 1 u = 931.5 MeV c⁻².
• Write nuclear reactions correctly in nuclear notation, conserving A and Z.
• Interpret the binding‑energy‑per‑nucleon curve and relate it to nuclear stability.
• Calculate the energy released (Q‑value) in a nuclear reaction.

1. Why We Know a Nucleus Exists – Rutherford Scattering

2. Simple Nuclear Model

A neutral atom consists of:

• Electrons – occupy the surrounding electron cloud (outside the nucleus).
• Nucleus – a compact core containing:

  • Protons (Z) – positively charged nucleons.
  • Neutrons (N) – neutral nucleons.

The total number of nucleons is the nucleon number (also called the mass number, A):

\[

A = Z + N

\]

In Cambridge terminology “mass number” and “nucleon number” are interchangeable, but the symbol A always represents the total number of nucleons.

3. Mass Defect

The mass of a bound nucleus is slightly less than the sum of the masses of its constituent free nucleons. This shortfall is the mass defect (Δm):

\[

\Delta m = \bigl(Z\,mp + N\,mn\bigr) - m_{\text{nucleus}}

\]

• Z – proton number (atomic number).
• N – neutron number (N = A − Z).
• m_p and m_n – masses of a free proton and neutron.
• m_nucleus – measured atomic mass of the neutral atom (electron masses are negligible for the nucleus).

4. From Mass Defect to Binding Energy

Einstein’s relation \$E=mc^{2}\$ links the mass defect to the energy required to separate the nucleus into its nucleons:

\[

E_{\text{b}} = \Delta m\,c^{2}

\]

Using the convenient conversion

\[

1\;\text{u} = 931.5\;\text{MeV}\,c^{-2}

\]

the binding energy (in MeV) is

\[

E_{\text{b}}(\text{MeV}) = \Delta m(\text{u}) \times 931.5

\]

5. Binding Energy per Nucleon

\[

\frac{E{\text{b}}}{A} = \frac{E{\text{b}}}{Z+N}

\]

• Plotting \$E_{\text{b}}/A\$ against \$A\$ yields the characteristic “binding‑energy curve”.
• The curve peaks near \$A\approx56\$ (e.g. \$^{56}\text{Fe}\$), indicating maximal stability.
• Light nuclei (\$A<56\$) release energy by fusion; heavy nuclei (\$A>56\$) release energy by fission.

6. Nuclear Notation and Writing Nuclear Equations

A nucleus is written as

\[

\prescript{A}{Z}{\text{X}}

\]

• Superscript \$A\$ – nucleon number (total nucleons).
• Subscript \$Z\$ – proton number.
• Letter \$X\$ – chemical symbol.

In any nuclear reaction both \$A\$ and \$Z\$ must be conserved:

\[

\prescript{A}{Z}{\text{X}} + \prescript{a}{z}{\text{y}} \;\rightarrow\; \prescript{A'}{Z'}{\text{X'}} + \prescript{b}{b}{\text{z}}

\]

Common Types of Decay

Fusion and Fission Examples

• Fusion (light nuclei):
  

  \[

  \prescript{2}{1}{\text{H}} + \prescript{3}{1}{\text{H}} \;\rightarrow\; \prescript{4}{2}{\text{He}} + \prescript{1}{0}{\text{n}}

  \]

• Fission (heavy nuclei):
  

  \[

  \prescript{235}{92}{\text{U}} + \prescript{1}{0}{\text{n}} \;\rightarrow\; \prescript{141}{56}{\text{Ba}} + \prescript{92}{36}{\text{Kr}} + 3\,\prescript{1}{0}{\text{n}}

  \]

7. Energy Released – Q‑value

The net energy change of a nuclear reaction is the Q‑value:

\[

Q = \bigl(m{\text{reactants}} - m{\text{products}}\bigr)c^{2}

= \bigl(m{\text{reactants}} - m{\text{products}}\bigr)\times 931.5\;\text{MeV}

\]

• If \$Q>0\$, the reaction is exothermic (energy released).
• If \$Q<0\$, the reaction is endothermic (energy must be supplied).

8. Sample Calculation – Binding Energy of \$^{4}\text{He}\$

1. Given masses (u): \$m{\alpha}=4.002603\$, \$m{p}=1.007276\$, \$m_{n}=1.008665\$.
2. Total mass of separate nucleons:

   \[

   m{\text{total}} = 2m{p}+2m_{n}=2(1.007276)+2(1.008665)=4.031882\;\text{u}

   \]

3. Mass defect:

   \[

   \Delta m = m{\text{total}}-m{\alpha}=4.031882-4.002603=0.029279\;\text{u}

   \]

4. Binding energy:

   \[

   E_{\text{b}} = 0.029279\times931.5 = 27.2\;\text{MeV}

   \]

5. Binding energy per nucleon:

   \[

   \frac{E_{\text{b}}}{A}= \frac{27.2}{4}=6.8\;\text{MeV nucleon}^{-1}

   \]

9. Table of Typical Binding Energies

10. Practice Questions

1. Binding energy of \$^{12}\text{C}\$:
   

   Given \$m{p}=1.007276\;\text{u}\$, \$m{n}=1.008665\;\text{u}\$, \$m{^{12}\text{C}}=12.000000\;\text{u}\$, calculate Δm, \$E{\text{b}}\$, and \$E_{\text{b}}/A\$.

2. Beta‑plus decay of \$^{11}\text{C}\$:
   

   Write the nuclear equation.

3. Q‑value for the fusion of two deuterium nuclei:
   

   \[

   \prescript{2}{1}{\text{H}} + \prescript{2}{1}{\text{H}} \rightarrow \prescript{3}{2}{\text{He}} + \prescript{1}{0}{\text{n}}

   \]
   

   Use \$m{^{2}\text{H}}=2.014102\;\text{u}\$, \$m{^{3}\text{He}}=3.016029\;\text{u}\$, \$m_{n}=1.008665\;\text{u}\$. State whether the reaction is exothermic.

4. Explain why \$^{4}\text{He}\$ cannot undergo fission to release energy.

11. Summary

The mass defect quantifies the loss of mass when nucleons bind together. Converting this loss to energy via \$E=mc^{2}\$ yields the nuclear binding energy, a direct measure of nuclear stability. The binding‑energy‑per‑nucleon curve peaks near \$^{56}\text{Fe}\$, explaining why light nuclei release energy by fusion and heavy nuclei by fission. Mastery of nuclear notation, the conservation of nucleon number (A) and proton number (Z), and Q‑value calculations enables clear description and quantitative analysis of all nuclear reactions required by the Cambridge A‑Level syllabus.