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Nuclear Physics - Radioactivity

Grade 12A LevelPhysics

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

🔑Concepts

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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.

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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.

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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: ZAX→Z−2A−4Y+24He^A_Z X \rightarrow ^{A-4}_{Z-2} Y + ^4_2\text{He}.

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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 νˉe\bar{\nu}_e are emitted. The atomic number increases by 11. Equation: ZAX→Z+1AY+−10e+νˉe^A_Z X \rightarrow ^A_{Z+1} Y + ^0_{-1}e + \bar{\nu}_e.

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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.

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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}).

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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.

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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.

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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=N0e−λtN = N_0 e^{-\lambda t}

A=λNA = \lambda N

λ=ln⁡(2)T1/2≈0.693T1/2\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 (800→400→200→100800 \rightarrow 400 \rightarrow 200 \rightarrow 100).

Problem 2:

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

Solution:

According to the conservation of nucleon number: 226=A+4⇒A=222226 = A + 4 \Rightarrow A = 222. According to the conservation of proton number: 88=Z+2⇒Z=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.