Cambridge Syllabus Notes

Objective

Students will be able to use the unified atomic mass unit (u) as a unit of mass, convert between u, kg and MeV, and apply these concepts to calculations involving atoms, nuclei and radiation, in full accordance with Cambridge AS & A‑Level Physics 9702 syllabus requirements 11.1 & 11.2.

1. Nuclear Notation and Nuclides

1.1 Symbols and numbers

Mass number (A): total number of protons + neutrons in the nucleus.
Atomic number (Z): number of protons; defines the element.
Nuclide symbol: \(\,^{A}_{Z}\text{X}\,\) where X is the chemical symbol, the superscript is \(A\) and the subscript is \(Z\). Example: \(\,^{238}_{92}\text{U}\,\).
Isotopes: nuclides with the same \(Z\) but different \(A\). They have identical chemical properties but different nuclear masses.

1.2 Conservation rules in nuclear reactions

Nucleon number (A) is always conserved: the sum of mass numbers before a reaction equals the sum after.
Charge (Z) is always conserved: the sum of atomic numbers before equals the sum after.
These rules apply to all nuclear processes, including α‑, β‑, γ‑decay, electron capture and positron emission.

1.3 Common reaction formats

α‑decay: \(\,^{A}_{Z}\text{X}\;\rightarrow\;^{A-4}_{Z-2}\text{Y}+^{4}_{2}\alpha\)
β⁻‑decay: \(\,^{A}_{Z}\text{X}\;\rightarrow\;^{A}_{Z+1}\text{Y}+e^{-}+\bar{u}_{e}\)
β⁺‑decay (positron emission): \(\,^{A}_{Z}\text{X}\;\rightarrow\;^{A}_{Z-1}\text{Y}+e^{+}+u_{e}\)
Electron capture: \(\,^{A}_{Z}\text{X}+e^{-}\;\rightarrow\;^{A}_{Z-1}\text{Y}+u_{e}\)
γ‑emission: \(\,^{A}_{Z}\text{X}^{*}\;\rightarrow\;^{A}_{Z}\text{X}+\gamma\)

2. The Unified Atomic Mass Unit (u)

2.1 Definition

The unified atomic mass unit, symbol , is defined as one twelfth of the mass of a neutral carbon‑12 atom:

\[ 1\ \text{u} = \frac{1}{12}\,m_{\text{C}^{12}} \]

2.2 Value in kilograms

\[ 1\ \text{u}=1.66053906660\times10^{-27}\ \text{kg} \]

2.3 Energy equivalence

\[ 1\ \text{u}=931.494\ \text{MeV}\;c^{-2} \] Thus a mass of \(m\) u corresponds to an energy \(E=m\times931.494\) MeV.

2.4 Why use the u?

Atomic and nuclear masses are of order \(10^{-27}\) kg; kilograms are unwieldy.
One u is close to the mass of a single nucleon, so the unit directly reflects nuclear composition.
The u provides a convenient bridge between mass and energy via \(E=mc^{2}\).

3. Relating Mass Number, Atomic Mass and Mass in u

3.1 Approximation

For most nuclides the atomic mass (in u) is close to the mass number:

\[ m \approx A\ \text{u} \]

Deviations arise from the mass‑defect caused by nuclear binding energy.

3.2 Example – Carbon‑12

By definition, \(\,^{12}_{6}\text{C}\) has a mass of exactly 12 u:

\[ 12\ \text{u}=12\times1.66053906660\times10^{-27}\ \text{kg}=1.9926468799\times10^{-26}\ \text{kg} \]

4. Converting Between u, kg and MeV

4.1 u ↔ kg

\[ m\ (\text{kg}) = m\ (\text{u})\times1.66053906660\times10^{-27}\ \text{kg u}^{-1} \]

4.2 u ↔ MeV (energy equivalent)

\[ E\ (\text{MeV}) = m\ (\text{u})\times931.494\ \text{MeV u}^{-1} \]

5. Types of Radiation

Radiation	Symbol	Charge	Mass (u)	Typical kinetic energy (MeV)	Decay equation
Alpha	\(\alpha\)	+2 e	4.002603 u	4–9	\(\,^{A}_{Z}\text{X}\rightarrow^{A-4}_{Z-2}\text{Y}+^{4}_{2}\alpha\)
Beta‑minus	\(\beta^{-}\)	− e	0.0005486 u (electron)	0.01–10	\(\,^{A}_{Z}\text{X}\rightarrow^{A}_{Z+1}\text{Y}+e^{-}+\bar{u}_{e}\)
Beta‑plus (positron)	\(\beta^{+}\)	+ e	0.0005486 u (positron)	0.01–5	\(\,^{A}_{Z}\text{X}\rightarrow^{A}_{Z-1}\text{Y}+e^{+}+u_{e}\)
Electron capture	\(EC\)	0	0 u (the captured electron’s mass is included in the atomic mass of the parent)	–	\(\,^{A}_{Z}\text{X}+e^{-}\rightarrow^{A}_{Z-1}\text{Y}+u_{e}\)
Gamma	\(\gamma\)	neutral	0 u (photon)	0.01–10+	\(\,^{A}_{Z}\text{X}^{*}\rightarrow^{A}_{Z}\text{X}+\gamma\)

5.1 Antiparticles and neutrinos

Positron (\(e^{+}\)): antiparticle of the electron, same mass, opposite charge.
Neutrino (\(u_{e}\)) and antineutrino (\(\bar{u}_{e}\)): electrically neutral, extremely small (practically zero) rest mass; they carry away excess energy and momentum in β‑decay.

6. Fundamental Particles (Syllabus 11.2)

The Standard Model groups particles into two families:

Leptons (6): electron (\(e^{-}\)), muon (\(\mu^{-}\)), tau (\(\tau^{-}\)) and their associated neutrinos (\(u_{e},u_{\mu},u_{\tau}\)). All have spin ½ and no colour charge.
Quarks (6): up (u), down (d), charm (c), strange (s), top (t), bottom (b). Quarks also have spin ½ but carry a colour charge and combine to form hadrons.

Hadrons are colour‑neutral composites of quarks:

Baryons – three quarks (e.g. proton \(uud\), neutron \(udd\)).
Mesons – a quark–antiquark pair (e.g. \(\pi^{+}=u\bar{d}\)).

β‑decay can be described at the quark level as a change of flavour mediated by the weak interaction:

\[ d \;\rightarrow\; u + e^{-} + \bar{u}_{e}\qquad u \;\rightarrow\; d + e^{+} + u_{e} \]

7. Mass Defect and Nuclear Binding Energy

7.1 Mass defect

The difference between the sum of the masses of the constituent nucleons (and electrons, when atomic masses are used) and the actual mass of the nuclide:

\[ \Delta m = \bigl(Z\,m_{p}+N\,m_{n}+Z\,m_{e}\bigr)-M_{\text{nuclide}} \]

7.2 Binding energy

\[ E_{\text{b}} = \Delta m\,c^{2}= \Delta m \times 931.494\ \text{MeV u}^{-1} \]

Binding energy per nucleon (\(E_{\text{b}}/A\)) is a useful indicator of nuclear stability.

8. Q‑Value Calculations

8.1 General formula

\[ Q = \bigl(\text{mass of reactants} - \text{mass of products}\bigr)\,c^{2} = \bigl(\Delta m\ \text{in u}\bigr)\times931.494\ \text{MeV} \]

8.2 Example – α‑decay of \(\,^{238}_{92}\text{U}\)

\[ ^{238}_{92}\text{U}\;\rightarrow\;^{234}_{90}\text{Th}+^{4}_{2}\alpha+Q \] \[ \begin{aligned} m(^{238}\text{U}) &= 238.050788\ \text{u}\\ m(^{234}\text{Th}) &= 234.043601\ \text{u}\\ m(^{4}\text{He}) &= 4.002603\ \text{u} \end{aligned} \] \[ \Delta m = 238.050788-(234.043601+4.002603)=0.004584\ \text{u} \] \[ Q = 0.004584\ \text{u}\times931.494\ \text{MeV/u}\approx4.27\ \text{MeV} \]

8.3 Example – β⁻‑decay of \(\,^{14}_{6}\text{C}\)

\[ ^{14}_{6}\text{C}\;\rightarrow\;^{14}_{7}\text{N}+e^{-}+\bar{u}_{e} \] \[ \begin{aligned} m(^{14}\text{C}) &= 14.003242\ \text{u}\\ m(^{14}\text{N}) &= 14.003074\ \text{u}\\ m(e^{-}) &= 0.0005486\ \text{u} \end{aligned} \] \[ \Delta m = 14.003242-(14.003074+0.0005486)= -0.0003806\ \text{u} \] \[ Q = |\Delta m|\times931.494\ \text{MeV/u}\approx0.355\ \text{MeV} \]

The negative sign simply indicates that the kinetic energy of the emitted electron and antineutrino equals the Q‑value.

8.4 Example – β⁺‑decay (positron emission) of \(\,^{22}_{11}\text{Na}\)

\[ ^{22}_{11}\text{Na}\;\rightarrow\;^{22}_{10}\text{Ne}+e^{+}+u_{e} \] \[ \begin{aligned} m(^{22}\text{Na}) &= 21.994436\ \text{u}\\ m(^{22}\text{Ne}) &= 21.991385\ \text{u}\\ m(e^{+}) &= 0.0005486\ \text{u} \end{aligned} \] \[ \Delta m = 21.994436-(21.991385+0.0005486)=0.002502\ \text{u} \] \[ Q = 0.002502\ \text{u}\times931.494\ \text{MeV/u}\approx2.33\ \text{MeV} \]

9. Summary Table of Common Nuclides

Nuclide	Symbol \(\,^{A}_{Z}\text{X}\,\)	Mass number (A)	Atomic mass (u)	Mass (kg)
Protium	\(^{1}_{1}\text{H}\)	1	1.007825	1.674 × 10⁻²⁷
Carbon‑12	\(^{12}_{6}\text{C}\)	12	12.000000	1.992 × 10⁻²⁶
Nitrogen‑14	\(^{14}_{7}\text{N}\)	14	14.003074	2.324 × 10⁻²⁶
Uranium‑235	\(^{235}_{92}\text{U}\)	235	235.043930	3.904 × 10⁻²⁵
Uranium‑238	\(^{238}_{92}\text{U}\)	238	238.050788	3.953 × 10⁻²⁵

10. Practice Questions

Calculate the mass in kilograms of a single \(^{56}_{26}\text{Fe}\) nucleus (atomic mass = 55.9349 u).
A β⁻ particle (electron) has a kinetic energy of 0.511 MeV. Using \(E=mc^{2}\), find its mass in u and compare with the accepted value 0.0005486 u.
Determine the Q‑value for the β⁺ decay \[ ^{22}_{11}\text{Na}\;\rightarrow\;^{22}_{10}\text{Ne}+e^{+}+u_{e} \] given:
- \(m(^{22}\text{Na}) = 21.994436\ \text{u}\)
- \(m(^{22}\text{Ne}) = 21.991385\ \text{u}\)
- \(m(e^{+}) = 0.0005486\ \text{u}\)
For the α‑decay of \(^{210}_{82}\text{Pb}\) to \(^{206}_{80}\text{Hg}\), calculate the Q‑value using the atomic masses:
- \(m(^{210}\text{Pb}) = 209.984188\ \text{u}\)
- \(m(^{206}\text{Hg}) = 205.977514\ \text{u}\)
- \(m(^{4}\text{He}) = 4.002603\ \text{u}\)
Explain why the mass of a neutral atom is slightly larger than the mass of its nucleus alone.

11. Suggested Diagram

Flow‑chart linking mass number (A), atomic mass (u) and mass (kg). Arrows display the conversion factors \(1\ \text{u}=1.6605\times10^{-27}\ \text{kg}\) and \(1\ \text{u}=931.5\ \text{MeV c}^{-2}\). A side box shows the nuclide notation \(\,^{A}_{Z}\text{X}\,\) and the conservation of \(A\) and \(Z\) in a generic nuclear reaction.

use the unified atomic mass unit (u) as a unit of mass