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Nuclear and Quantum Physics - Radioactive Decay

Grade 11IBPhysics

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

🔑Concepts

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Radioactive decay is a random and spontaneous process by which an unstable atomic nucleus loses energy by emitting radiation. The rate of decay is proportional to the number of unstable nuclei present: A=λNA = \lambda N.

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α\alpha (Alpha) Decay: Emission of a helium nucleus 24He^{4}_{2}\text{He}. The parent nuclide's mass number AA decreases by 44 and the atomic number ZZ decreases by 22.

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β−\beta^{-} (Beta-minus) Decay: A neutron decays into a proton, an electron −10e^{0}_{-1}e (beta particle), and an electron antineutrino νˉe\bar{\nu}_e. ZZ increases by 11, AA remains constant.

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β+\beta^{+} (Beta-plus) Decay: A proton decays into a neutron, a positron +10e^{0}_{+1}e, and an electron neutrino νe\nu_e. ZZ decreases by 11, AA remains constant.

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γ\gamma (Gamma) Decay: Emission of high-energy photons when a nucleus transitions from an excited state to a lower energy state. AA and ZZ remain unchanged.

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Half-life (T1/2T_{1/2}): The time taken for the number of radioactive nuclei (or the activity) in a sample to decrease to half of its initial value.

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Activity (AA): The number of decays per unit time, measured in Becquerels (1 Bq=1 decay per second1\text{ Bq} = 1\text{ decay per second}).

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Background Radiation: Constant ionizing radiation present in the environment from natural sources (cosmic rays, radon gas) and artificial sources (medical X-rays).

📐Formulae

A=λNA = \lambda N

T1/2=ln⁡2λ≈0.693λT_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda}

N=N0(12)nN = N_0 \left(\frac{1}{2}\right)^n

n=tT1/2n = \frac{t}{T_{1/2}}

A=A0e−λtA = A_0 e^{-\lambda t}

💡Examples

Problem 1:

A radioactive isotope of Iodine-131 has a half-life of 8.08.0 days. If a sample initially contains 1.6×10201.6 \times 10^{20} undecayed nuclei, calculate the activity of the sample after 2424 days. Express the answer in Bq\text{Bq}.

Solution:

  1. Find the number of half-lives: n=tT1/2=248.0=3n = \frac{t}{T_{1/2}} = \frac{24}{8.0} = 3.
  2. Find the remaining nuclei NN: N=N0(12)3=1.6×1020×0.125=2.0×1019N = N_0 \left(\frac{1}{2}\right)^3 = 1.6 \times 10^{20} \times 0.125 = 2.0 \times 10^{19}.
  3. Calculate the decay constant λ\lambda: λ=ln⁡2T1/2=0.6938.0×24×3600≈1.003×10−6 s−1\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{8.0 \times 24 \times 3600} \approx 1.003 \times 10^{-6} \text{ s}^{-1}.
  4. Calculate Activity AA: A=λN=(1.003×10−6)×(2.0×1019)≈2.0×1013 BqA = \lambda N = (1.003 \times 10^{-6}) \times (2.0 \times 10^{19}) \approx 2.0 \times 10^{13} \text{ Bq}.

Explanation:

First, we determine how many half-lives have passed to find the remaining number of nuclei. Then, because Activity is defined as A=λNA = \lambda N, we convert the half-life into seconds to find the decay constant in s−1\text{s}^{-1} before calculating the final activity.

Problem 2:

Complete the following decay equation: 92238U→90234Th+X^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + X. Identify particle XX.

Solution:

92238U→90234Th+24α^{238}_{92}\text{U} \rightarrow ^{234}_{90}\text{Th} + ^{4}_{2}\alpha (or 24He^{4}_{2}\text{He}). XX is an Alpha particle.

Explanation:

In any nuclear reaction, the total mass number AA and atomic number ZZ must be conserved. 238=234+4238 = 234 + 4 and 92=90+292 = 90 + 2. The particle with A=4A=4 and Z=2Z=2 is the Helium nucleus, known as an alpha particle.

Radioactive Decay - Revision Notes & Key Formulas | IB Grade 11 Physics