Cambridge AS & A Level9702

Nuclear physics

Physics 9702 Chapter Notes

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Nuclear physics - Mass defect and nuclear binding energyNuclear physics - Radioactive decay
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1. Mass-Energy Equivalence

At the heart of nuclear physics is one of the most famous equations in science: E = mc². Proposed by Albert Einstein, it reveals a fundamental relationship between energy (E) and mass (m). It states that mass is a form of energy and can be converted into other forms of energy, and vice versa. The constant c² (the speed of light squared) is a huge number, meaning a tiny amount of mass can be converted into a vast amount of energy. This is the principle behind the energy released in nuclear reactions. In nuclear physics, masses are often measured in unified atomic mass units (u), where 1 u is defined as 1/12th the mass of a carbon-12 atom. It's useful to know the energy equivalent of 1 u.

E = mc²

1 u = 1.6605 × 10⁻²⁷ kg

Energy equivalent of 1 u ≈ 931.5 MeV

Key term

Mass-energy equivalence: The principle that mass and energy are interchangeable properties of a system, related by the equation E = mc², where c is the speed of light.

Examiner insight

Examiners expect you to recall and apply the E = mc² equation. Marks are often awarded for correctly converting units (e.g., grams to kilograms) before substitution.

Common pitfall

When using E = mc², a common mistake is forgetting to convert the mass to the standard unit of kilograms (kg) before calculating the energy in joules.

Worked example 13 marks

Calculate the energy released, in joules, when 0.50 g of mass is converted completely into energy. (Speed of light, c = 3.00 × 10⁸ m s⁻¹)

  1. 1

    Step 1: State the mass-energy equivalence formula. E = mc²

  2. 2

    Step 2: Convert the mass from grams to kilograms. m = 0.50 g = 0.50 × 10⁻³ kg

  3. 3

    Step 3: Substitute the values for mass and the speed of light into the equation. E = (0.50 × 10⁻³ kg) × (3.00 × 10⁸ m s⁻¹)²

  4. 4

    Step 4: Calculate the result. E = (0.50 × 10⁻³) × (9.00 × 10¹⁶) = 4.5 × 10¹³ J

Recap

  • Mass and energy are interchangeable, as described by E = mc².
  • A small amount of mass can be converted into a very large amount of energy.
  • The unified atomic mass unit (u) is a common unit for mass in nuclear physics.
  • The energy equivalent of 1 u is approximately 931.5 MeV.

Quick check

  1. What does 'c' represent in the equation E = mc²?1 mark
  2. How much energy, in MeV, is equivalent to a mass of 2 u?1 mark

2. Mass Defect and Binding Energy

If you were to weigh a nucleus (like helium-4) and then weigh its individual components (2 protons and 2 neutrons) separately, you'd find something strange: the separate components weigh more than the assembled nucleus. This difference in mass is called the mass defect (Δm). So where does the 'missing' mass go? It is converted into energy when the nucleus is formed, according to E = mc². This energy is called the binding energy (E_B). The binding energy is the energy that holds the nucleus together. It is also, therefore, the minimum energy required to completely separate a nucleus into its constituent protons and neutrons.

Mass Defect (Δm) = (Total mass of separate nucleons) - (Mass of nucleus)

Binding Energy (E_B) = Δm c²

Key term

Mass Defect: The difference between the mass of an intact nucleus and the sum of the masses of its separate constituent protons and neutrons.

Examiner insight

Show your working clearly. State the formula, show the substitution for the mass of the constituents, the calculation of the mass defect, and the final conversion to energy. Each step can earn marks.

Common pitfall

A frequent error is subtracting the masses in the wrong order when calculating the mass defect. Remember, the separate parts are always heavier than the whole nucleus.

Worked example 14 marks

A nucleus of helium-4 (⁴₂He) has a mass of 4.002603 u. A proton has a mass of 1.007276 u and a neutron has a mass of 1.008665 u. Calculate the binding energy of the helium-4 nucleus in MeV.

  1. 1

    Step 1: Find the number of protons and neutrons. Helium-4 has 2 protons and 4 - 2 = 2 neutrons.

  2. 2

    Step 2: Calculate the total mass of the separate nucleons. Mass = (2 × mass of proton) + (2 × mass of neutron) = (2 × 1.007276u) + (2 × 1.008665u) = 2.014552 u + 2.017330 u = 4.031882 u.

  3. 3

    Step 3: Calculate the mass defect (Δm). Δm = (Mass of nucleons) - (Mass of nucleus) = 4.031882 u - 4.002603 u = 0.029279 u.

  4. 4

    Step 4: Convert the mass defect into binding energy in MeV. We know 1 u is equivalent to 931.5 MeV. E_B = 0.029279 u × 931.5 MeV/u = 27.27 MeV.

Recap

  • The mass of a nucleus is always less than the total mass of its individual protons and neutrons.
  • This difference in mass is called the mass defect.
  • The mass defect is converted into binding energy, which holds the nucleus together.
  • Binding energy is calculated using E_B = Δm c².
  • Binding energy is the energy released when a nucleus is formed, or the energy needed to break it apart.

Quick check

  1. What is binding energy?2 marks

3. Binding Energy Per Nucleon

While a large binding energy means a nucleus is tightly held together, it doesn't tell the whole story. A very large nucleus will naturally have a large total binding energy. A better measure of nuclear stability is the binding energy per nucleon, which is the total binding energy divided by the number of nucleons (protons + neutrons) in the nucleus. The higher the binding energy per nucleon, the more stable the nucleus. A graph of binding energy per nucleon against nucleon number (A) shows a distinct pattern. It rises sharply for light nuclei, peaks at a nucleon number of around 56 (Iron), and then slowly decreases for heavier nuclei. This curve is key to understanding energy release in nuclear reactions.

Binding Energy per Nucleon = Binding Energy / Nucleon Number

BE/A = E_B / A

Key term

Binding Energy per Nucleon: The average energy required to remove one nucleon from a nucleus, which serves as a measure of the nucleus's stability.

Examiner insight

When explaining nuclear energy release, always frame your answer in terms of an increase in the binding energy per nucleon of the products compared to the reactants.

Common pitfall

Confusing total binding energy with binding energy per nucleon. A heavy nucleus like uranium has a very large total binding energy, but its binding energy per nucleon is lower than that of iron, making it less stable.

Worked example 12 marks

The binding energy of a Uranium-235 nucleus is 1784 MeV. Calculate its binding energy per nucleon.

  1. 1

    Step 1: Identify the total binding energy and the nucleon number. E_B = 1784 MeV. The nucleon number (A) for Uranium-235 is 235.

  2. 2

    Step 2: State the formula for binding energy per nucleon. BE/A = E_B / A.

  3. 3

    Step 3: Substitute the values and calculate. BE/A = 1784 MeV / 235 = 7.59 MeV per nucleon.

Worked example 23 marks

Using the result from the previous worked example (Helium-4 binding energy = 27.27 MeV), calculate its binding energy per nucleon and state whether it is more or less stable per nucleon than Uranium-235.

  1. 1

    Step 1: Calculate the binding energy per nucleon for Helium-4. The nucleon number (A) is 4. BE/A = 27.27 MeV / 4 = 6.82 MeV per nucleon.

  2. 2

    Step 2: Compare the values. Uranium-235 has BE/A = 7.59 MeV/nucleon. Helium-4 has BE/A = 6.82 MeV/nucleon.

  3. 3

    Step 3: Conclude. Since 7.59 > 6.82, a nucleon in a Uranium-235 nucleus is, on average, more tightly bound than one in a Helium-4 nucleus. However, both are less stable than nuclei near the peak of the curve (e.g. Iron, with BE/A ≈ 8.8 MeV).

Recap

  • Binding energy per nucleon is a measure of nuclear stability.
  • The higher the binding energy per nucleon, the more stable the nucleus.
  • The graph of binding energy per nucleon peaks at nucleon number A ≈ 56 (Iron).
  • Nuclei with very low or very high nucleon numbers are less stable than those in the middle.
  • Energy is released when reactions cause the products to have a higher binding energy per nucleon.

Quick check

  1. Which element has the most stable nuclei in the universe?1 mark

4. Nuclear Fission and Fusion

The binding energy curve explains why two types of nuclear reactions release enormous amounts of energy: fission and fusion.

Nuclear Fission: This is the process where a heavy, unstable nucleus (like Uranium-235) splits into two or more smaller, lighter nuclei. This happens because the heavy nucleus is on the right-hand side of the binding energy peak. When it splits, the resulting smaller nuclei are closer to the peak and thus have a higher binding energy per nucleon. This increase in binding energy per nucleon is released as energy. Fission is the process used in nuclear power plants.

Nuclear Fusion: This is the process where two light nuclei (like isotopes of hydrogen) combine to form a single, heavier nucleus. This happens because the light nuclei are on the left-hand, rising part of the binding energy curve. When they fuse, the resulting nucleus is further up the curve and has a higher binding energy per nucleon. Again, this increase is released as energy. Fusion is the process that powers the Sun and other stars.

Energy Released = (Final Total Mass - Initial Total Mass) c² = Δm c²

Energy Released = (Final Total Binding Energy) - (Initial Total Binding Energy)

Key term

Nuclear Fission: A nuclear reaction in which a heavy nucleus splits into smaller, lighter nuclei, releasing a large amount of energy and typically several neutrons.

Fun fact

A single pellet of uranium fuel, about the size of a gummy bear, can produce the same amount of energy as a ton of coal, 149 gallons of oil or 17,000 cubic feet of natural gas.

Worked example 13 marks

In a nuclear reactor, a uranium-235 nucleus absorbs a neutron and undergoes fission: ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3¹₀n. Explain, with reference to binding energy, why this reaction releases energy.

  1. 1

    Step 1: Identify the reactants and products. The reactant is a heavy nucleus (Uranium-235). The products are lighter nuclei (Barium-141 and Krypton-92).

  2. 2

    Step 2: Relate this to the binding energy per nucleon curve. Heavy nuclei like uranium have a lower binding energy per nucleon than medium-sized nuclei like barium and krypton.

  3. 3

    Step 3: Formulate the explanation. The products of the fission reaction (barium and krypton) have a higher binding energy per nucleon than the original uranium nucleus. This means the product nuclei are more stable. The difference in total binding energy between the products and the reactant is released as energy.

Recap

  • Fission is the splitting of a heavy nucleus into lighter ones.
  • Fusion is the joining of light nuclei to form a heavier one.
  • Both fission and fusion can release energy.
  • Energy is released because the products have a higher binding energy per nucleon than the reactants.
  • Fission moves from the heavy end of the BE curve towards the peak.
  • Fusion moves from the light end of the BE curve towards the peak.

Quick check

  1. Which process, fission or fusion, powers the Sun?1 mark
  2. Why does splitting a heavy nucleus release energy?2 marks

5. Random and Spontaneous Decay

Radioactive decay, the process by which an unstable nucleus transforms by emitting radiation, is governed by two key principles: it is spontaneous and random.

Spontaneous: This means the decay of a nucleus is not affected by any external factors. You cannot make a nucleus decay by heating it, crushing it, or putting it in a chemical reaction. The decay happens on its own, independent of its environment.

Random: This means it is impossible to predict when a particular nucleus will decay. If you have two identical unstable nuclei, one might decay in the next second, while the other might last for a million years. We have no way of knowing which will be which. However, for a very large number of nuclei (like in any macroscopic sample), the overall rate of decay is predictable. The evidence for this randomness comes from observing the count rate from a radioactive source with a Geiger counter. The clicks or readings will not be perfectly regular; they will fluctuate about an average value, showing that the decays are happening at random moments.

Key term

Random Decay: The principle that it is impossible to predict which nucleus in a sample will decay next, or when any individual nucleus will decay.

Examiner insight

A common question asks for the evidence for the random nature of decay. The expected answer is the observation of fluctuations in count rate when measured with a detector like a Geiger-Müller tube.

Worked example 13 marks

A student measures the count rate from a radioactive source over a period of 60 seconds. They notice the number of counts recorded in each 5-second interval is not the same. Explain this observation.

  1. 1

    Step 1: State the nature of radioactive decay. Radioactive decay is a random process.

  2. 2

    Step 2: Explain what 'random' means in this context. This means we cannot predict when the next nucleus will decay.

  3. 3

    Step 3: Link the concept to the observation. Because the decays occur at random, unpredictable moments, the number of decays occurring in any short time interval will vary. This leads to fluctuations in the measured count rate.

Recap

  • Radioactive decay is a spontaneous process, unaffected by external conditions.
  • Radioactive decay is a random process, meaning individual decays are unpredictable.
  • Evidence for the random nature of decay comes from fluctuations in the measured count rate.
  • While individual decays are random, the behavior of a large sample is statistically predictable.

Quick check

  1. State the meaning of the term 'spontaneous' in the context of radioactive decay.1 mark

6. Activity, Decay Constant and Half-Life

To describe the decay of a large number of nuclei, we use three key quantities.

Activity (A): This is the rate at which nuclei in a sample are decaying. It's measured in becquerels (Bq), where 1 Bq = 1 decay per second.

Decay Constant (λ): This is the probability that a single, individual nucleus will decay in a unit of time. It has units of s⁻¹, min⁻¹, or year⁻¹, etc. A large decay constant means a high probability of decay, so the substance decays quickly.

These two are related to the number of undecayed nuclei (N) in the sample by the equation A = λN.

Half-life (t₁/₂): This is a more intuitive measure. It is the average time taken for half of the undecayed nuclei in a sample to decay. It is also the time taken for the activity of the sample to fall to half its initial value. A short half-life means a substance is highly radioactive but will become safe quickly. A long half-life means it's less radioactive but will remain a hazard for a long time. Half-life and the decay constant are inversely related.

A = λN

λ = 0.693 / t₁/₂

t₁/₂ = 0.693 / λ

ln(2) ≈ 0.693

Key term

Half-life (t₁/₂): The average time taken for the number of undecayed nuclei in a sample of a radioactive isotope to be reduced by half.

Common pitfall

Ensure that the time units for the decay constant (e.g., s⁻¹) and the half-life (e.g., s) are consistent when using the formula λ = 0.693 / t₁/₂.

Worked example 13 marks

A sample of cobalt-60 has a half-life of 5.27 years. Calculate its decay constant in s⁻¹.

  1. 1

    Step 1: State the relationship between half-life and decay constant. λ = 0.693 / t₁/₂.

  2. 2

    Step 2: Convert the half-life into seconds. t₁/₂ = 5.27 years × 365.25 days/year × 24 hours/day × 3600 s/hour = 1.663 × 10⁸ s.

  3. 3

    Step 3: Substitute the half-life in seconds into the formula. λ = 0.693 / (1.663 × 10⁸s) = 4.17 × 10⁻⁹ s⁻¹.

Worked example 22 marks

A sample contains 5.0 × 10¹⁵ undecayed nuclei of a radioisotope with a decay constant λ = 1.4 × 10⁻¹¹ s⁻¹. Calculate the activity of the sample.

  1. 1

    Step 1: State the formula for activity. A = λN.

  2. 2

    Step 2: Substitute the given values. A = (1.4 × 10⁻¹¹ s⁻¹) × (5.0 × 10¹⁵).

  3. 3

    Step 3: Calculate the activity. A = 7.0 × 10⁴ Bq.

Recap

  • Activity (A) is the rate of decay, measured in Bq.
  • Decay constant (λ) is the probability of decay per unit time for one nucleus.
  • Activity is proportional to the number of undecayed nuclei: A = λN.
  • Half-life (t₁/₂) is the time for half the nuclei to decay.
  • Half-life and decay constant are related by t₁/₂ = 0.693 / λ.

Quick check

  1. If a sample has a half-life of 10 days, what fraction of the original sample will be left after 30 days?2 marks

7. The Mathematics of Exponential Decay

The random nature of decay at the individual level leads to a predictable, smooth, exponential decrease for a large population of nuclei. The number of undecayed nuclei, the activity, and the measured count rate all decrease exponentially with time. This behaviour is described by a single, powerful equation: x = x₀e⁻ˡᵗ. In this equation, x₀ is the initial value of the quantity (at time t=0), x is the value at time t, λ is the decay constant, and 'e' is the base of the natural logarithm (approximately 2.718). This equation is the cornerstone of all calculations involving radioactive decay over time, from medical imaging to carbon dating.

N = N₀e⁻ˡᵗ

A = A₀e⁻ˡᵗ

R = R₀e⁻ˡᵗ

Key term

Exponential Decay: A process in which a quantity decreases at a rate proportional to its current value, described by the equation x = x₀e⁻ˡᵗ.

Examiner insight

Examiners often test the rearrangement of the exponential decay equation to find the time, t. Being comfortable with using natural logarithms (ln) is essential for these questions.

Common pitfall

When measuring count rate (R) to use in decay equations, it is crucial to first subtract the background count rate from the measured value to get the true count rate from the source.

Worked example 14 marks

A sample of Iodine-131 has a half-life of 8.0 days. If a sample initially has an activity of 640 MBq, what will its activity be after 24 days?

  1. 1

    Step 1: Calculate the decay constant (λ). λ = 0.693 / t₁/₂ = 0.693 / 8.0 days = 0.086625 day⁻¹.

  2. 2

    Step 2: State the exponential decay formula for activity. A = A₀e⁻ˡᵗ.

  3. 3

    Step 3: Substitute the known values. A₀ = 640 MBq, λ = 0.086625 day⁻¹, t = 24 days.

  4. 4

    Step 4: Calculate the final activity. A = 640 × e⁻⁽⁰.⁰⁸⁶⁶²⁵ × ²⁴⁾ = 640 × e⁻².⁰⁷⁹ = 640 × 0.125 = 80 MBq.

  5. 5

    Alternative Step (using half-lives): 24 days is 24/8 = 3 half-lives. After 1 half-life, A = 320 MBq. After 2 half-lives, A = 160 MBq. After 3 half-lives, A = 80 MBq.

Worked example 24 marks

An ancient wooden axe handle has a carbon-14 activity of 0.125 Bq. A sample of modern wood of the same mass has an activity of 1.0 Bq. If the half-life of carbon-14 is 5730 years, estimate the age of the axe handle.

  1. 1

    Step 1: Calculate the decay constant (λ). λ = 0.693 / t₁/₂ = 0.693 / 5730 years = 1.209 × 10⁻⁴ year⁻¹.

  2. 2

    Step 2: State the decay equation A = A₀e⁻ˡᵗ and rearrange for t. A/A₀ = e⁻ˡᵗ => ln(A/A₀) = -λt => t = -ln(A/A₀) / λ.

  3. 3

    Step 3: Substitute the values. A = 0.125 Bq, A₀ = 1.0 Bq, λ = 1.209 × 10⁻⁴ year⁻¹.

  4. 4

    Step 4: Calculate the time t. t = -ln(0.125 / 1.0) / (1.209 × 10⁻⁴) = -ln(0.125) / (1.209 × 10⁻⁴) = -(-2.079) / (1.209 × 10⁻⁴) = 17190 years.

  5. 5

    Alternative Step (using half-lives): The activity has dropped to 1/8 of the original (0.125/1.0). 1/8 = (1/2)³. This means 3 half-lives have passed. Age = 3 × 5730 years = 17190 years.

Recap

  • Radioactive decay follows an exponential pattern.
  • The equation x = x₀e⁻ˡᵗ describes the decay of N, A, or R over time.
  • x₀ is the initial quantity, and x is the quantity at time t.
  • This equation can be used to find the age of artefacts (radiometric dating).
  • Always ensure λ and t are in consistent time units.

Quick check

  1. In the equation N = N₀e⁻ˡᵗ, what does N₀ represent?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A nucleus of plutonium-239 (²³⁹₉₄Pu) decays by emitting an alpha particle (⁴₂He). Write a balanced nuclear equation for this decay and identify the daughter nucleus.3 marks
  2. Define binding energy and explain how it relates to the mass defect of a nucleus.3 marks
  3. The mass of a lithium-7 nucleus (⁷₃Li) is 7.01435 u. The mass of a proton is 1.00728 u and the mass of a neutron is 1.00866 u. Calculate the binding energy per nucleon for lithium-7 in MeV. (1 u = 931.5 MeV/c²).5 marks
  4. Sketch the graph of binding energy per nucleon against nucleon number. Use your graph to explain why energy is released in both nuclear fission and nuclear fusion.4 marks
  5. A particular fusion reaction is: ²₁H + ³₁H → ⁴₂He + ¹₀n. The masses are: ²₁H = 2.01410 u, ³₁H = 3.01605 u, ⁴₂He = 4.00260 u, ¹₀n = 1.00866 u. Calculate the energy released in this reaction in joules. (1 u = 1.66 × 10⁻²⁷ kg, c = 3.00 × 10⁸ m s⁻¹).4 marks
  6. Explain what is meant by the terms 'spontaneous' and 'random' with reference to radioactive decay.2 marks
  7. Radon-222 has a half-life of 3.8 days. A sample of radon-222 has an initial activity of 200 Bq. Calculate the activity of the sample after 11.4 days.3 marks
  8. A radioactive source has a half-life of 25 minutes. At a certain time, its measured count rate is 460 counts per minute. The background count rate is 60 counts per minute. Calculate the time, in minutes, at which the count rate from the source will have fallen to 50 counts per minute.5 marks
  9. A sample of a radioactive isotope has an activity of 3.2 × 10⁹ Bq. The half-life of the isotope is 12 hours. Calculate the initial number of undecayed nuclei in the sample.4 marks
  10. A hospital uses a sample of technetium-99m for a diagnostic scan. The sample is 'fresh' and has an activity of 800 MBq at 9:00 am. The half-life of technetium-99m is 6.0 hours. A patient scan requires an activity of at least 250 MBq. What is the latest time of day the sample can be used?5 marks

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