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Nuclear and Quantum Physics - Radioactive Decay

Grade 12IBPhysics

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

🔑Concepts

Radioactive decay is a random and spontaneous process. The probability of decay per unit time is called the decay constant λ\lambda.

The activity AA of a sample 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}).

The number of undecayed nuclei NN decreases exponentially over time according to the law N=N0eλtN = N_0 e^{-\lambda t}.

Half-life T1/2T_{1/2} is the time taken for the number of undecayed nuclei (or the activity) to reduce to half of its initial value. It is related to the decay constant by T1/2=ln2λT_{1/2} = \frac{\ln 2}{\lambda}.

In α\alpha decay, a nucleus emits a 24He^4_2\text{He} nucleus. In β\beta^- decay, a neutron transforms into a proton, emitting an electron (ee^-) and an anti-neutrino (νˉe\bar{\nu}_e). In β+\beta^+ decay, a proton transforms into a neutron, emitting a positron (e+e^+) and a neutrino (νe\nu_e).

Gamma emission (γ\gamma) occurs when a nucleus transitions from an excited state to a lower energy state, emitting high-energy photons without changing the atomic or mass number.

📐Formulae

N=N0eλtN = N_0 e^{-\lambda t}

A=λNA = \lambda N

A=A0eλtA = A_0 e^{-\lambda t}

T1/2=ln2λT_{1/2} = \frac{\ln 2}{\lambda}

λ=0.693T1/2\lambda = \frac{0.693}{T_{1/2}}

💡Examples

Problem 1:

A radioactive isotope has a half-life of 1414 days. If the initial activity of a sample is 6.4×106 Bq6.4 \times 10^6 \text{ Bq}, calculate the activity after 4242 days.

Solution:

n=tT1/2=4214=3 half-livesn = \frac{t}{T_{1/2}} = \frac{42}{14} = 3 \text{ half-lives} A=A02n=6.4×10623=6.4×1068=8.0×105 BqA = \frac{A_0}{2^n} = \frac{6.4 \times 10^6}{2^3} = \frac{6.4 \times 10^6}{8} = 8.0 \times 10^5 \text{ Bq}

Explanation:

Since the time elapsed is an exact multiple of the half-life, we can calculate the number of half-lives nn and then divide the initial activity by 2n2^n.

Problem 2:

Find the decay constant λ\lambda for a substance with a half-life of 5.0×103 s5.0 \times 10^3 \text{ s}.

Solution:

λ=ln2T1/2=0.6935.0×103=1.39×104 s1\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{5.0 \times 10^3} = 1.39 \times 10^{-4} \text{ s}^{-1}

Explanation:

The decay constant is inversely proportional to the half-life and represents the probability of a nucleus decaying per unit time.

Radioactive Decay - Revision Notes & Key Formulas | IB Grade 12 Physics