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Nuclear Physics - Radioactivity

Grade 12IGCSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Radioactivity is the spontaneous and random emission of radiation from the unstable nucleus of an atom. It is unaffected by chemical or physical conditions such as temperature or pressure.

The structure of the nucleus is represented by the notation ZAX^A_Z X, where AA is the nucleon (mass) number and ZZ is the proton (atomic) number.

Alpha (α\alpha) decay: The emission of a helium nucleus 24He^4_2\text{He}. The mass number decreases by 44 and the atomic number decreases by 22. Equation: ZAXZ2A4Y+24He^A_Z X \rightarrow ^{A-4}_{Z-2} Y + ^4_2\text{He}.

Beta-minus (β\beta^-) decay: A neutron in the nucleus decays into a proton and an electron. The electron 10e^0_{-1}e and an antineutrino uˉe\bar{ u}_e are emitted. The atomic number increases by 11. Equation: ZAXZ+1AY+10e+uˉe^A_Z X \rightarrow ^A_{Z+1} Y + ^0_{-1}e + \bar{ u}_e.

Gamma (γ\gamma) radiation: High-energy electromagnetic waves emitted to allow the nucleus to reach a lower energy state. There is no change to AA or ZZ.

Activity (AA) is the rate at which a source decays, measured in Becquerels (1 Bq=1 decay per second1\text{ Bq} = 1\text{ decay per second}).

Half-life (T1/2T_{1/2}) is the average time taken for half of the radioactive nuclei in a sample to decay, or for the activity to fall to half of its initial value.

Nuclear Fission: The splitting of a heavy nucleus (e.g., 92235U^{235}_{92}\text{U}) into two smaller daughter nuclei and neutrons, releasing a large amount of energy.

Nuclear Fusion: The process where two light nuclei (e.g., isotopes of Hydrogen like 12H^2_1\text{H} and 13H^3_1\text{H}) combine to form a heavier nucleus, releasing energy.

📐Formulae

A=Z+NA = Z + N

N=N0eλtN = N_0 e^{-\lambda t}

A=λNA = \lambda N

λ=ln(2)T1/20.693T1/2\lambda = \frac{\ln(2)}{T_{1/2}} \approx \frac{0.693}{T_{1/2}}

A=A0(12)nA = A_0 \left(\frac{1}{2}\right)^n

💡Examples

Problem 1:

A sample of radioactive material has an initial activity of 800 Bq800\text{ Bq}. If the half-life of the isotope is 66 hours, calculate the activity of the sample after 1818 hours.

Solution:

Step 1: Calculate the number of half-lives (nn): n=total timehalf-life=18 h6 h=3n = \frac{\text{total time}}{\text{half-life}} = \frac{18\text{ h}}{6\text{ h}} = 3. Step 2: Use the activity formula: A=A0×(12)n=800×(12)3=800×18=100 BqA = A_0 \times (\frac{1}{2})^n = 800 \times (\frac{1}{2})^3 = 800 \times \frac{1}{8} = 100\text{ Bq}.

Explanation:

Since 1818 hours represents exactly three half-life periods, the activity halves three times (800400200100800 \rightarrow 400 \rightarrow 200 \rightarrow 100).

Problem 2:

Complete the nuclear equation for the alpha decay of Radium-226: 88226RaZARn+24α^{226}_{88}\text{Ra} \rightarrow ^A_Z\text{Rn} + ^4_2\alpha.

Solution:

According to the conservation of nucleon number: 226=A+4A=222226 = A + 4 \Rightarrow A = 222. According to the conservation of proton number: 88=Z+2Z=8688 = Z + 2 \Rightarrow Z = 86. The resulting nucleus is Radon: 86222Rn^{222}_{86}\text{Rn}.

Explanation:

In α\alpha decay, the parent nucleus loses two protons and two neutrons, reducing the mass number by 44 and the atomic number by 22.

Radioactivity - Revision Notes & Key Formulas | IGCSE Grade 12 Physics