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Nuclear and Quantum Physics - Radioactivity

Grade 12IBPhysics

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

🔑Concepts

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Radioactive decay is a random and spontaneous process at the single-nucleus level, but follows a predictable exponential trend for large populations.

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α\alpha-decay involves the emission of a helium nucleus 24He^{4}_{2}\text{He}. The parent nuclide's atomic number ZZ decreases by 22 and mass number AA decreases by 44.

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β−\beta^{-}-decay occurs when a neutron turns into a proton, emitting an electron (e−e^{-}) and an electron antineutrino (νˉe\bar{\nu}_e). ZZ increases by 11.

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β+\beta^{+}-decay occurs when a proton turns into a neutron, emitting a positron (e+e^{+}) and an electron neutrino (νe\nu_e). ZZ decreases by 11.

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γ\gamma-decay is the 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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The decay constant λ\lambda represents the probability of decay per unit time for a single nucleus.

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Activity (AA) is the rate of decay, defined as A=−dNdt=λNA = -\frac{dN}{dt} = \lambda N, 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 time taken for half the radioactive nuclei in a sample to decay, or for the activity to halve.

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Mass defect (Δm\Delta m) is the difference between the mass of a nucleus and the sum of the masses of its individual nucleons. Binding energy is E=Δmc2E = \Delta m c^2.

📐Formulae

N=N0e−λtN = N_0 e^{-\lambda t}

A=A0e−λtA = A_0 e^{-\lambda t}

A=λNA = \lambda N

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

E=Δmc2E = \Delta m c^2

Number of nuclei N=n×NA=mM×NA\text{Number of nuclei } N = n \times N_A = \frac{m}{M} \times N_A

💡Examples

Problem 1:

A sample of Carbon-14 (614C^{14}_{6}\text{C}) has an initial activity of 800 Bq800\text{ Bq}. Given that the half-life of Carbon-14 is 5730 years5730\text{ years}, calculate the activity of the sample after 17190 years17190\text{ years}.

Solution:

  1. Identify the number of half-lives elapsed: n=tT1/2=171905730=3n = \frac{t}{T_{1/2}} = \frac{17190}{5730} = 3.
  2. Use the activity formula: A=A0(12)nA = A_0 \left(\frac{1}{2}\right)^n.
  3. Substitute values: A=800×(12)3=800×18=100 BqA = 800 \times \left(\frac{1}{2}\right)^3 = 800 \times \frac{1}{8} = 100\text{ Bq}.

Explanation:

Activity decreases by half every half-life. Since 1719017190 years corresponds to exactly 33 half-lives, the activity is halved three times: 800→400→200→100800 \rightarrow 400 \rightarrow 200 \rightarrow 100.

Problem 2:

Calculate the decay constant λ\lambda for a radioactive isotope that has a half-life of 20 minutes20\text{ minutes}. Express your answer in s−1s^{-1}.

Solution:

  1. Convert half-life to seconds: T1/2=20×60=1200 sT_{1/2} = 20 \times 60 = 1200\text{ s}.
  2. Use the formula: λ=ln⁡2T1/2\lambda = \frac{\ln 2}{T_{1/2}}.
  3. Calculate: λ=0.6931200≈5.78×10−4 s−1\lambda = \frac{0.693}{1200} \approx 5.78 \times 10^{-4}\text{ s}^{-1}.

Explanation:

The decay constant is inversely proportional to the half-life. Standard SI units for decay constants are s−1s^{-1}.

Problem 3:

A sample contains 2.5×10202.5 \times 10^{20} nuclei of an isotope with a decay constant λ=4.0×10−6 s−1\lambda = 4.0 \times 10^{-6}\text{ s}^{-1}. Determine the initial activity of the sample.

Solution:

  1. Use the relationship between activity and number of nuclei: A=λNA = \lambda N.
  2. Substitute the given values: A=(4.0×10−6 s−1)×(2.5×1020)A = (4.0 \times 10^{-6}\text{ s}^{-1}) \times (2.5 \times 10^{20}).
  3. Calculate: A=1.0×1015 BqA = 1.0 \times 10^{15}\text{ Bq}.

Explanation:

Activity is directly proportional to the number of undecayed nuclei present in the sample.

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