3.8.1.3
Finding N from mass using the Avogadro constant
Radioactive Decay & Half-Life — AQA A-Level Physics
Key Definition
Avogadro constant (Nₐ) — The number of atoms in one mole of a substance. Nₐ = 6.02 × 10²³ mol⁻¹.
$$N = \frac{m \times N_A}{M}$$
- $N$: number of nuclei
- $m$: mass of sample (g)
- $N_A$: Avogadro constantThe number of particles in one mole of a substance. N_A = 6.02 x 10²³ mol⁻¹. = 6.02 × 10²³ mol⁻¹
- $M$: molar mass (g mol⁻¹)
- The molar mass in g mol⁻¹ is numerically equal to the nucleon number A.
- For example, 241 g of Am-241 contains 6.02 × 10²³ atoms.
Worked Example
A smoke detector contains 5.1 μg of Am-241 (molar mass 241) with activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second. 5.9 × 10⁵ Bq. The original activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second. was 6.1 × 10⁵ Bq. Find (a) the number of nuclei, (b) the decay constantThe probability of decay of a nucleus per unit time. Measured in s⁻¹., and (c) the age.
Show Solution
1
(a) Number of nuclei
$$N = \frac{5.1 \times 10^{-6} \times 6.02 \times 10^{23}}{241} = 1.27 \times 10^{16}$$
2
(b) Decay constantThe probability of decay of a nucleus per unit time. Measured in s⁻¹.
$$\lambda = \frac{A}{N} = \frac{5.9 \times 10^5}{1.27 \times 10^{16}} = 4.65 \times 10^{-11} \text{ s}^{-1}$$
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(c) Age from exponential decay
$$A = A_0 e^{-\lambda t}$$
$$t = -\frac{1}{\lambda} \ln\left(\frac{A}{A_0}\right) = -\frac{1}{4.65 \times 10^{-11}} \ln\left(\frac{5.9 \times 10^5}{6.1 \times 10^5}\right)$$
$$t = 7.17 \times 10^8 \text{ s} = \frac{7.17 \times 10^8}{365 \times 24 \times 3600} = 22.7 \text{ years}$$Answer
(a) N = 1.27 × 10¹⁶ nuclei, (b) λ = 4.65 × 10⁻¹¹ s⁻¹, (c) Age ≈ 22.7 years