Physics notes, chapter 6

Nuclear Physics

Textbook Physics Form 5 (DLP), pages 198–221

What this chapter covers

Describe radioactive decay as a random and spontaneous process, and write decay equations for alpha, beta and gamma decay (6.1.1)
Explain half-life and the uranium-238 decay series, and how they are used to estimate the age of rocks and the Earth (6.1.2)
Determine half-life from a decay curve and use N = (1/2)^n N0 to solve problems (6.1.3, 6.1.4)
Describe nuclear fission and nuclear fusion, and explain mass defect and E = mc² (6.2.1, 6.2.2)
Calculate the nuclear energy released in radioactive decay and nuclear reactions (6.2.3)
Explain how a nuclear reactor generates electricity, the chain reaction, and how the reaction is controlled safely (6.2.4)
Discuss nuclear energy as an alternative source of electrical energy (6.2.5)

6.1Radioactive decay

Radioactive decay is the process in which an unstable nucleus becomes more stable by emitting radiation. It is random (we cannot tell which nucleus decays next) and spontaneous (it is not affected by temperature, pressure or other physical conditions).

There are three kinds of radiation: alpha particles (α), beta particles (β) and gamma rays (γ). In a decay equation the nucleon number A (top) and the proton number Z (bottom) must each add up to the same total on both sides.

Decay	What is emitted	General equation	Example
Alpha	A helium nucleus (2 protons, 2 neutrons)	ᴬ_ZX → ᴬ⁻⁴_Z−2Y + ⁴₂He	²¹⁰₈₄Po → ²⁰⁶₈₂Pb + ⁴₂He
Beta	A fast electron; a neutron in the nucleus changes to a proton plus an electron (n → p + e)	ᴬ_ZX → ᴬ_Z+1Y + ⁰₋₁e	²⁴₁₁Na → ²⁴₁₂Mg + ⁰₋₁e
Gamma	A high-frequency electromagnetic wave with no mass and no charge; the nucleus loses excess energy	ᴬ_ZX → ᴬ_ZX + γ	⁶⁰₂₇Co* → ⁶⁰₂₇Co + γ

In beta decay the nucleon number stays the same but the proton number goes up by 1. In gamma decay neither number changes: only the energy does.

Half-life and the decay series

The half-life, T_½, is the time taken for the number of undecayed radioactive nuclei in a sample to fall to half of its initial value.

The activity of a sample (decays per second) is proportional to the number of undecayed nuclei, so the activity also halves every half-life. A shorter half-life means a faster decay and a steeper curve.

Decay curve: the fraction of nuclei left halves from 1 to 0.5, 0.25 and 0.125 after 1, 2 and 3 half-lives — Decay curve. Equal steps in time give equal factors of ½. The curve never reaches zero.

After n half-lives: N = (½)ⁿ N₀, where n = t ÷ T_½. The same formula works for activity: A = (½)ⁿ A₀.

Half-lives passed	0	1	2	3	4	5
Fraction left	1	½	¼	⅛	1/16	1/32

Worked example

A sample has an initial activity of 480 s⁻¹ and a half-life of 6 minutes. Find its activity after 30 minutes.

n = 30 ÷ 6 = 5. A = (½)⁵ × 480 = 480 ÷ 32 = 15 s⁻¹.

A new nucleus formed by decay may itself be unstable, so the decay continues in a decay series until a stable nucleus forms. In the uranium-238 series the end product is stable lead-206, through 8 α-decays and 6 β-decays.

Uranium-238 decay series from uranium-238 to lead-206 with the half-life of each step — The uranium-238 decay series. (Textbook Figure 6.5, p. 202)

Figure: Physics Form 5 (DLP) textbook, Figure 6.5, p. 202.

Dating rocks

Uranium-238 has a very long half-life (about 4.5 × 10⁹ years) and is trapped when a rock forms. By measuring the ratio of lead-206 to uranium-238 in a sample, geologists estimate how old the rock is. The higher the ratio, the older the rock. The same method gives the age of the Earth.

Worked example

A volcanic rock has argon-40 : potassium-40 = 3 : 1. There was no argon at the start. The half-life of potassium-40 is 1 250 million years. Estimate the age.

Ratio 1 : 1 after one half-life, 3 : 1 after two (¾ decayed, ¼ left). Age = 2 × 1 250 = 2 500 million years.

(a) Complete: ²²⁸₈₈Ra → A + ⁰₋₁e. (b) A source has a half-life of 5 days. What fraction is left after 15 days?

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(a) A has nucleon number 228 and proton number 89: ²²⁸₈₉Ac. (b) 15 ÷ 5 = 3 half-lives, so (½)³ = ⅛ is left.

6.2Nuclear energy

Nuclear fission

A heavy nucleus splits into two or more lighter nuclei and releases a large amount of energy.

²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3 ¹₀n + energy

Nuclear fusion

Small, light nuclei join to form a heavier nucleus and release a large amount of energy. It needs extremely high temperature and pressure, as in the core of the Sun.

²₁H + ³₁H → ⁴₂He + ¹₀n + energy

Mass defect and E = mc²

The mass defect, m, is the loss of mass in a nuclear reaction: total mass before minus total mass after. The energy released is E = mc², with m in kg and c = 3.00 × 10⁸ m s⁻¹.

One atomic mass unit is 1/12 of the mass of a carbon-12 atom: 1 amu = 1.66 × 10⁻²⁷ kg. Nuclear energy is also given in megaelectronvolts: 1 MeV = 1.6 × 10⁻¹³ J.

Worked example

Radium-226 (226.54 amu) decays to radon-222 (222.018 amu) and helium-4 (4.003 amu). Find the energy released.

m = 226.54 − (222.018 + 4.003) = 0.519 amu = 0.519 × 1.66 × 10⁻²⁷ = 8.62 × 10⁻²⁸ kg.

E = mc² = 8.62 × 10⁻²⁸ × (3.00 × 10⁸)² = 7.75 × 10⁻¹¹ J.

Convert the mass defect from amu to kg before using E = mc². The answer is then in joules.

Generating electricity in a nuclear reactor

In a reactor, fission of uranium-235 produces two daughter nuclei, three fast neutrons and energy. These neutrons strike other uranium nuclei, giving a chain reaction that continues by itself.

Nuclear power station: reactor, steam generator, turbine, generator, condenser and cooling tower — Reactor core heats water in the steam generator. The high-pressure steam turns the turbine and generator, then condenses. (Textbook Figure 6.12, p. 214)

Figure: Physics Form 5 (DLP) textbook, Figure 6.12, p. 214.

Energy changes: nuclear energy → heat energy → kinetic energy (turbine) → electrical energy (generator).

Part	Job
Uranium rods	Fuel for fission
Control rods (boron or cadmium)	Absorb excess neutrons to control the rate of reaction
Moderator (graphite or water)	Slows down fast neutrons so that fission can occur
Water coolant	Absorbs heat to make steam and controls the temperature
Thick concrete walls	Stop radiation leaking to the surroundings

Nuclear energy gives large power with a low carbon footprint, but it needs high cost, expert technology and careful radioactive waste management. Malaysia has a research reactor (TRIGA PUSPATI) but no nuclear power plant.

A fusion reaction has a mass defect of 0.018 amu. Find the energy released in joules.

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m = 0.018 × 1.66 × 10⁻²⁷ = 2.99 × 10⁻²⁹ kg. E = 2.99 × 10⁻²⁹ × (3.0 × 10⁸)² = 2.69 × 10⁻¹² J.