Radioactive Decay Calculator
Calculate remaining nuclei, activity, or decay constant using N = N₀·e^(−λt) and A = λN.
What Is Radioactive Decay?
Radioactive decay follows an exponential law: N(t) = N₀·e^(−λt), where N₀ is the initial number of radioactive nuclei, λ is the decay constant (s⁻¹), and t is time. The relationship between λ and half-life: λ = ln(2)/t½ = 0.693/t½. Activity A = λN (decays per second, measured in Becquerels).
The exponential form arises because each nucleus decays independently with constant probability λ per unit time. The differential equation dN/dt = −λN gives the exponential solution. This is a first-order process — the same mathematics governs capacitor discharge (Q = Q₀e^(−t/RC)), drug elimination, and Newton's law of cooling.
Mean lifetime (τ = 1/λ = t½/ln2) is the average time a nucleus survives before decaying. After one mean lifetime, N = N₀/e ≈ 36.8% remains (compared to 50% after one half-life). Radioactivity A = λN = (ln2/t½)N means a sample with more atoms but same t½ has proportionally higher activity.
Units: 1 Becquerel (Bq) = 1 decay per second. 1 Curie (Ci) = 3.7×10¹⁰ Bq. Absorbed dose: 1 Gray (Gy) = 1 J/kg. Effective dose: 1 Sievert (Sv) = dose in Gy × quality factor (1 for gamma, 20 for alpha). Annual background radiation ≈ 2–3 mSv.
Formula Reference Table
| Solve For | Formula | Notes |
|---|---|---|
| Decay law | N(t) = N₀·e^(−λt) | λ = decay constant (s⁻¹) |
| Activity | A = λ·N | Bq = decays/s |
| Half-life relation | λ = ln(2)/t½ = 0.693/t½ | t½ in same units as t |
| Mean lifetime | τ = 1/λ = t½/ln(2) | τ = 1.443 × t½ |
| Activity at time t | A(t) = A₀·e^(−λt) | Activity also decays exponentially |
| Becquerel | 1 Bq = 1 decay/s | 1 Ci = 3.7×10¹⁰ Bq |
3 Worked Examples
1 g of ancient wood has N = 5.97×10¹⁰ atoms of C-14. t½ = 5,730 yr = 1.808×10¹¹ s. Find activity.
- λ = 0.693/(1.808×10¹¹) = 3.83×10⁻¹² s⁻¹
- A = λN = 3.83×10⁻¹² × 5.97×10¹⁰ = 0.229 Bq
- Modern C-14 activity in 1 g wood ≈ 0.226 Bq — confirms this sample is approximately 'fresh' (current)
Patient receives 370 MBq of I-131 (t½ = 8.02 days). Activity after 24 days?
- λ = 0.693/(8.02×86400) = 1.0×10⁻⁶ s⁻¹
- t = 24 days = 2.074×10⁶ s
- A = 370×10⁶ × e^(−1.0×10⁻⁶ × 2.074×10⁶) = 370×10⁶ × e^(−2.074)
- A = 370 × 0.126 = 46.6 MBq
Carbon-14: t½ = 5,730 years. Find λ in s⁻¹.
- t½ = 5730 × 365.25 × 24 × 3600 = 1.807×10¹¹ s
- λ = ln(2)/t½ = 0.6931/1.807×10¹¹ = 3.836×10⁻¹² s⁻¹
- Mean lifetime τ = 1/λ = 2.608×10¹¹ s = 8,267 years
Real-World Applications
Common Mistakes to Avoid
λ must have units of inverse time (matching t). If t½ is in years, λ = ln2/t½ is in yr⁻¹. For A = λN in Bq (s⁻¹), convert λ to s⁻¹ and ensure N is in atoms (not grams).
N = number of radioactive atoms (not Bq). A = λN = decay rate in Bq. Doubling λ doubles A even with same N. Doubling N doubles A for same λ.
t½ = 0.693/λ; τ = 1/λ = 1.443×t½. After one mean lifetime: 36.8% remains. After one half-life: 50% remains.
N = N₀e^(−λt), not N₀(e^λ)^t or N₀e^(+λt). The exponent must be negative for decay.
To get activity in Bq: λ must be in s⁻¹. C-14 t½ = 5,730 yr = 1.807×10¹¹ s; λ = 3.84×10⁻¹² s⁻¹.
Frequently Asked Questions
Related Physics Calculators
Formula Explorer connections
Interpretation: This relationship connects quantized energy, wavelength, probability, nuclear mass or radioactive change. Assumption: Use the correct particle, quantum state, nuclide and energy units. Idealized potentials, nonrelativistic motion or single decay channels may be assumed.