The radioactive isotope ⁸²Sr has a half-life of 25.4 days. A sample containing this isotope has an initial activity at t = 0 of 4.50 × 108 Bq. Calculate the number of nuclei that will decay in the time interval between t₁ = 34.0 hours and t₂ = 70.0 hours.

Radioactive Decay Calculation for ⁸²Sr

Calculate the Number of Nuclei Decayed in a Radioactive Sample

Question:
The radioactive isotope ⁸²Sr has a half-life of 25.4 days. A sample containing this isotope has an initial activity at t = 0 of 4.50 × 108 Bq. Calculate the number of nuclei that will decay in the time interval between t₁ = 34.0 hours and t₂ = 70.0 hours.

🧮 Step-by-Step Solution:

Step 1: Convert hours to days
  • t₁ = 34.0 hours = 34 / 24 = 17/12 ≈ 1.4167 days
  • t₂ = 70.0 hours = 70 / 24 = 35/12 ≈ 2.9167 days
Step 2: Calculate the decay constant (λ)
λ = ln(2) / half-life = 0.69314718 / 25.4 ≈ 0.02728925 day⁻¹
Step 3: Use the decay formula
Activity at time t: A(t) = A₀ × e-λt

At t₁:
A(t₁) = 4.5 × 10⁸ × e−(0.02728925 × 17/12) ≈ 432935091.21 Bq

At t₂:
A(t₂) = 4.5 × 10⁸ × e−(0.02728925 × 35/12) ≈ 415571190.65 Bq
Step 4: Find the difference in activity to get decayed nuclei
Decayed = A(t₁) − A(t₂)
Decayed = 432935091.21 − 415571190.65 ≈ 17363900.55
Final Answer:
The number of nuclei that decayed between t₁ = 34.0 hours and t₂ = 70.0 hours is approximately:
1.74 × 10⁷ nuclei

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