⚛️ Nuclei Count & Half-Life
☢️ Activity & Decay Constant
🪵 Age Finder (Isotope Ratio)

Radioactivity Fundamentals

Radioactive decay is a spontaneous and random process by which unstable nuclei lose energy by emitting radiation. Because it is random, we use statistical laws to describe the behavior of large numbers of nuclei.

N = N₀ e-λt

1. Decay Constant (λ)

The decay constant represents the probability per unit time that a specific nucleus will decay. It is a fundamental property of the isotope and is unaffected by external factors like temperature or pressure.

λ = ln(2) / t½ ≈ 0.693 / t½

2. Activity (A)

Activity is defined as the rate of decay, or the number of nuclei decaying per unit time. It is measured in Becquerels (Bq), where 1 Bq = 1 decay per second.

A = λN = A₀ e-λt
Exam Tip: Be careful with units! If you are calculating activity in Bq, you MUST ensure λ is in s⁻¹. Often half-life is given in years or days, so conversion is the most common place learners lose marks.

Deep Dive: Worked Examples

✅ Example 1: Activity in Becquerels

A sample contains 1.0 × 10¹² atoms of Carbon-14 (t½ = 5730 years). Calculate its activity in Bq.

Step 1: Convert half-life to seconds t½ = 5730 × 365.25 × 24 × 3600 ≈ 1.81 × 10¹¹ s
Step 2: Calculate λ λ = ln(2) / (1.81 × 10¹¹) = 3.83 × 10⁻¹² s⁻¹
Step 3: Solve for A = λN A = (3.83 × 10⁻¹²) × (1.0 × 10¹²) = 3.83 Bq

✅ Example 2: Determining Age (Carbon Dating)

A wooden artifact has an activity of 4.0 Bq per gram. Fresh wood has 16.0 Bq per gram. Find the age if t½ = 5730y.

Step 1: Check the ratio A / A₀ = 4.0 / 16.0 = 0.25 (or 1/4)
Step 2: Number of half-lives (1/2)ⁿ = 1/4 → n = 2 half-lives
Step 3: Total Age Age = 2 × 5730 = 11,460 years

✅ Example 3: Fraction Remaining

What fraction of a Radon-222 sample remains after 12 days? The half-life is 3.8 days.

Step 1: Calculate λ λ = 0.693 / 3.8 = 0.1824 days⁻¹
Step 2: Apply the decay equation N/N₀ = e^(-λt) = e^(-0.1824 × 12)
Step 3: Solve N/N₀ = e^(-2.1888) ≈ 0.112 (or 11.2%)

✅ Example 4: Molar Activity

Calculate the number of grams in a Cobolt-60 source (t½ = 5.27y) that has an activity of 10 GBq.

Step 1: Convert units A = 10 × 10⁹ Bq, λ = 4.17 × 10⁻⁹ s⁻¹ (converted from years)
Step 2: Find N N = A / λ = 10⁹ / 4.17⁻⁹ ≈ 2.40 × 10¹⁸ nuclei
Step 3: Convert N to mass m = (N × 60) / (6.02 × 10²³) ≈ 0.24 μg

✅ Example 5: Logarithmic Graphing

A student plots ln(Activity) vs Time. The gradient is -0.045 s⁻¹. What is the half-life?

Step 1: Interpret the gradient ln(A) = ln(A₀) - λt. This is y = mx + c. Gradient = -λ.
Step 2: Identify λ λ = 0.045 s⁻¹
Step 3: Find t½ t½ = ln(2) / 0.045 ≈ 15.4 seconds