Radioactive decay for OCR A-level Physics
This page covers the following topics:
1. Radioactive decay
2. Constant decay probability
3. Calculating activity
A nucleus is said to be stable if it has a certain amount of neutrons depending on the number of protons it has. If there are too many or too few neutrons, the nucleus is said to be unstable and will decay by emitting radiation. The same number of protons and neutrons are needed for nuclei of elements of fewer protons to be stable. The greater the number of protons, the more neutrons are needed for the nucleus to be stable. Radioactive decay occurs randomly and each decay is independent of the rest, therefore predictions about when an atom will decay cannot be made.
The decay constant, given by λ, is the probability of a given nucleus decaying per second. It is measured in s⁻¹. It is calculated using the following formula: t₁/₂ = ln2/λ, where t₁/₂ is the half time of the isotope. The half time is defined as the time taken for the mass of an isotope to be halved or the time taken for its activity to be halved.
The greater the time that has passed, the smaller the number of nuclei present, therefore the decay of the sample of the object will decrease. The number of decaying nuclei is directly proportional to the original number of them present. This can be expressed through the following formula: A = A₀exp(−λt), where A₀ is the initial activity and λ is the decay constant. Activity can also be calculated using the following formula: A = λN, where N is the number of nuclei present.
Given that the number of nuclei present in a radioactive sample with decay constant of 3.5 × 10⁻⁶ s⁻¹ after 20 minutes is 6.8 × 10⁸, calculate the initial activity of the sample.
Using A = λN, A = 3.5 × 10⁻⁶ s⁻¹ × 6.8 × 10⁸ = 2380 Bq. 20 minutes = 1200 seconds. Using A = A₀exp(−λt), 2380 Bq = A₀ × exp(-3.5 × 10⁻⁶ s⁻¹ × 1200 s), so A₀ = 2390 Bq (to the nearest integer).
Define the half-time of an isotope.
Half time is defined as the time taken for the mass of an isotope to be halved or the time taken for its activity to be halved.
Calculate the initial activity of a sample with a decay constant of 4.7 × 10⁻⁶ s⁻¹. The number of nuclei present in the sample after 45 minutes is found to be 4.2 × 10⁸.
Using A = λN, A = 4.7 × 10⁻⁶ s⁻¹ × 4.2 × 10⁸ = 1974 Bq. 45 minutes = 2700 seconds. Using A = A₀exp(−λt), 1974 Bq = A₀ × exp(-4.7 × 10⁻⁶ s⁻¹ × 2700 s), so A₀ = 1999 Bq (to the nearest integer).
The initial number of atoms present in an isotope are 7.2 × 10⁹. Given that its decay constant is 1.85 × 10⁻⁶ s⁻¹, calculate the initial activity of the isotope.
Using A = λN, A = 1.85 × 10⁻⁶ s⁻¹ × 7.2 × 10⁹ = 13320 Bq.
What is the relationship between the number of decaying nuclei and the original number of them present?
The number of decaying nuclei is directly proportional to the original number of them present.
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