Carbon-14 after one half-life
- Solve for
- Amount left
- Isotope
- Carbon-14
- Starting amount N₀
- 100
- Elapsed time
- 5700 yr
- Amount left N
- 50
- Half-lives elapsed
- 1
Nguồn đối chiếu: Definition of half-life (t½ = 5700 y, NNDC)
Half-life calculator: amount left, starting amount, elapsed time or half-life from N = N₀·2^(−t/t½), with carbon-14 and other isotope presets.
Cập nhật Ví dụ đã kiểm tra: 8
After 1 half-life (5,700 years), 50% of Carbon-14 remains: 50 of 100.
Radioactive material decays exponentially: after each half-life t½ half of the remaining nuclei are left, so N = N₀·2^(−t/t½), or N₀e^(−λt) with decay constant λ = ln 2/t½. Enter any three of starting amount, amount left, elapsed time and half-life to get the fourth, along with the fraction remaining, the number of half-lives and the mean lifetime 1/λ.
The default, 100 g of carbon-14 after 5,700 years, leaves 50 g: one half-life. Radiocarbon dating runs the same equation backwards, so a sample with 25% of the carbon-14 in living tissue is two half-lives, or 11,400 years, old. After 30 days, 7.49% of an iodine-131 dose remains.
Half-lives come from the NNDC NuDat 3 database. Amounts can be in any unit (grams, atoms, becquerels or percent) as long as both use the same one. Published radiocarbon ages use the Libby half-life of 5,568 years by convention and are then calibrated, so they differ from this raw calculation.
Nguồn đối chiếu: Definition of half-life (t½ = 5700 y, NNDC)
Nguồn đối chiếu: Two half-lives: 2 × 5700 y
Nguồn đối chiếu: Python 3.8 decimal: 100 × 2^(−30/8.0252) = 7.4934578
Nguồn đối chiếu: Python 3.8 decimal: t½ = 30 ln2 / ln 8 = 10 min = 0.0069444 d
t½ = t·ln 2/ln(N₀/N), from a starting amount N₀ and the amount N left after time t. If a count rate falls from 1,000 to 125 in 30 minutes, N₀/N = 8 = 2³, so three half-lives have passed and t½ = 10 minutes. The logarithm handles any ratio: 1,000 falling to 300 in 30 minutes gives t½ = 17.3 minutes.
12.5%. Each half-life halves what remains: 50% after one, 25% after two, 12.5% after three and 6.25% after four. After n half-lives the fraction left is (1/2)ⁿ, so falling to 1% takes 6.64 half-lives and falling below 0.1% takes 10 (0.098% remains).
5,700 years in the evaluated nuclear data published by the NNDC; older textbooks give 5,730 years. Radiocarbon dating reaches back about 50,000 years, close to nine half-lives, after which less than 0.3% of the original carbon-14 remains, too little to measure reliably.
Mean lifetime τ is the average time a nucleus survives before it decays: τ = 1/λ = t½/ln 2 ≈ 1.443 t½. For carbon-14 that is 5,700/0.693 = 8,223 years. After one mean lifetime 1/e, or 36.8%, of the sample remains, compared with 50% after one half-life.
λ = ln 2/t½, the probability per unit time that a given nucleus decays. It links amount to activity through A = λN, in becquerels when N counts atoms and λ is per second. Carbon-14's λ is 3.85 × 10⁻¹² per second, so 1 g of pure carbon-14, 4.3 × 10²² atoms, has an activity of about 1.66 × 10¹¹ Bq.
Độ chính xác phụ thuộc vào dữ liệu nhập và giả định của phương pháp. Phép tính thập phân dùng 50 chữ số có nghĩa, nhưng ước lượng, phương pháp số và dữ liệu nguồn có thể kém chính xác hơn; làm tròn khi hiển thị không loại bỏ các giới hạn đó. Ví dụ có lời giải đã đối chiếu với nguồn độc lập: 8. Ví dụ, “Carbon-14 after one half-life” được kiểm tra bằng Definition of half-life (t½ = 5700 y, NNDC).
NNDC NuDat 3 (ENSDF evaluated half-lives), Brookhaven National Laboratory; OpenStax University Physics Volume 3, §10.3 Radioactive decay.
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