Why doesn't first-order half-life depend on the initial amount?

For a first-order process the rate is proportional to how much reactant (or radioactive nuclide) is present: r = k[A]. Solving the differential equation gives [A]_t = [A]_0 · exp(−k·t), so the ratio [A]_t/[A]_0 depends only on k·t — the [A]_0 cancels. The time for the ratio to fall to 1/2, t_½ = ln(2)/k, therefore involves only k. Zero- and second-order half-lives DO depend on [A]_0 (t_½ = [A]_0/(2k) for zero-order; t_½ = 1/(k[A]_0) for second-order), and that is one way to identify the order experimentally: if repeated measurements of t_½ change as [A]_0 changes, the reaction is not first-order.

What is the difference between activity and amount for radioactive decay?

The number of radioactive nuclei N is the *amount* of material; the activity A = −dN/dt = λ·N is the number of disintegrations per second (measured in becquerels, Bq, or curies, 1 Ci = 3.7 × 10¹⁰ Bq). Both decay with the same half-life because λ (the decay constant) is a constant of the nuclide. So A_t/A_0 = N_t/N_0 = exp(−λ·t) = (1/2)^(t/t_½). You can therefore work the arithmetic in whichever the problem gives you — a mass, a mole count, or an activity — and this calculator's math is identical.

How is radioactive decay related to first-order chemical kinetics?

Both are described by the same differential equation: the rate is proportional to the amount present. For chemistry that's r = k[A] with rate constant k in s⁻¹; for a radionuclide it is dN/dt = −λN with decay constant λ in s⁻¹. The two constants play identical mathematical roles, and t_½ = ln(2)/k = ln(2)/λ. The physical mechanism is different — quantum-mechanical tunneling out of a nuclear potential well vs. classical activation-energy barrier crossing — but the observable macroscopic law is the same.

When does the half-life formula not apply?

The clean t_½ = ln(2)/k relation is a first-order result. Zero-order and second-order integrated rate laws produce concentration-dependent half-lives (see the equation panel above). Complex mechanisms — reversible reactions near equilibrium, autocatalytic reactions, chain reactions, saturable enzyme kinetics — do not admit a single half-life at all: the apparent t_½ changes as the reaction progresses. If your reactant is not decaying exponentially, no single number of the form 'ln(2)/k' will describe it. For radioactive decay, the exponential law is exact (a first-order quantum-mechanical process); for chemistry, verify first-order behaviour by plotting ln[A] vs. t and checking that the line is straight.

Which value should I use for the carbon-14 half-life?

Radiocarbon dating uses two conventions. The 'Libby half-life' 5,568 y (Libby 1949) is embedded in the calibration curves and reported ¹⁴C ages by convention — you keep using it so historical results stay comparable. The 'Cambridge half-life' 5,730 ± 40 y (Godwin 1962; adopted widely in physics texts) is closer to the modern best measurement. NIST's evaluated 2015 value is 5,700 ± 30 y. For teaching and back-of-envelope work, use 5,700 or 5,730 y; for anything you'll compare to a published radiocarbon calibration, keep 5,568 y as the conventional half-life and report the age as 'conventional radiocarbon years BP'.

Half-Life Calculator — First-Order Kinetics & Decay

Half-life is the time it takes for half of a decaying substance — a radioactive nuclide, a first-order chemical reactant, a plasma-drug concentration — to disappear. For a first-order process the number is deceptively simple: t_½ = ln(2)/k, independent of how much you started with. That single-number characterisation is why chemists, physicists, and pharmacokineticists all reach for it.

The first-order integrated rate law

If species A disappears at a rate proportional to how much of it is present — the differential rate law −d[A]/dt = k[A], with k in s⁻¹ — separating variables and integrating from time 0 to t gives the integrated rate law

$[A]_t = [A]_0 \, e^{-k t}, \quad t_{1/2} = \dfrac{\ln 2}{k}$

Notice that [A]₀ drops out of t½. That is the key insight of first-order kinetics: the time it takes for the concentration to halve is the same no matter how far along the reaction is. From 1.0 M → 0.5 M takes as long as from 0.001 M → 0.0005 M.

When the initial amount does matter

For zero- and second-order reactions it does. The integrated forms and half-lives are:

Order	Integrated rate law	Half-life
Zero	[A]ₜ = [A]₀ − k t	t½ = [A]₀ / (2k)
First	[A]ₜ = [A]₀ · exp(−k t)	t½ = ln 2 / k
Second	1/[A]ₜ − 1/[A]₀ = k t	t½ = 1 / (k [A]₀)

Watch the units of k carefully — they carry the order information. First-order k has units of s⁻¹; zero-order has M·s⁻¹; second-order (single reactant, r = k[A]²) has M⁻¹·s⁻¹. If a textbook problem quotes k without units, it is almost always because the author is being sloppy — clarify before plugging in.

Radioactive decay is first-order

Every unstable nuclide decays with a rate proportional to how many nuclei are left, so radioactive decay is always first-order and the half-life is a fixed property of the nuclide. Common examples, all values from NNDC or NIST:

Nuclide	Half-life
¹⁴C (carbon-14)	5,700 ± 30 y (NIST) — 5,730 y (Cambridge)
⁹⁹ᵐTc (technetium)	6.0072 h
¹³¹I (iodine)	8.0252 d
²³⁸U (uranium)	4.468 × 10⁹ y
³H (tritium)	12.32 y

The “decay constant” λ is exactly the first-order rate constant k in nuclear physics notation. Activity (disintegrations per second) tracks the number of nuclei: A(t) = λ N(t) = A₀ · exp(−λ t). You can therefore work the math on either.

Worked example 1 — first-order chemistry

A reactant A decomposes by first-order kinetics with k = 2.3 × 10⁻³ s⁻¹ and [A]₀ = 0.500 M. What is [A] after 300 s?

Worked example — [A] at 300 s

k: 2.3 × 10⁻³ s⁻¹
[A]₀: 0.500 M
t: 300 s

Formula
$[A]_{t} = [A]_{0} e^{- k t}$
Substitute
$[A]_{t} = 0.500 \cdot e^{- (2.3 \times 1 0^{- 3}) (300)}$
Note
k·t = 0.69, one natural log below zero — so [A] falls to about half of [A]₀. That's not a coincidence: 300 s is very close to the half-life, t½ = ln(2)/k = 301.4 s.
Result
$[A]_{t} = 0.500 \cdot e^{- 0.69} = 0.251 M$

Worked example 2 — carbon-14 dating

An organic sample retains 25 % of the ¹⁴C activity of a modern reference. How old is it?

Worked example — 25 % of modern ¹⁴C

f = A_t/A_0: 0.25 (25 %)
t½ (¹⁴C, NIST 2015): 5,700 y

Formula
$t = t_{1/2} \cdot \frac{ln ( 1/ f )}{ln 2}$
Substitute
$t = 5, 700 \cdot \frac{ln 4}{ln 2} = 5, 700 \cdot 2$
Note
Two half-lives elapsed: 100 % → 50 % → 25 %. The answer is exactly 2·t½, independent of the specific t½ value you pick — dating with the Cambridge value 5,730 y gives 11,460 y; with the Libby value 5,568 y gives 11,136 y. The convention matters when comparing published radiocarbon results.
Result
$t = 11, 400 y$

Choosing the right half-life for ¹⁴C

The literature has three values for the ¹⁴C half-life in circulation, and mixing them is a common source of error:

5,568 y (“Libby half-life”, 1949) — Willard Libby’s original measurement, still the conventional half-life used to report radiocarbon ages so historical dates stay comparable across the calibration curves (IntCal, SHCal).
5,730 ± 40 y (“Cambridge half-life”, Godwin 1962) — a revised measurement widely adopted in physics and geology textbooks.
5,700 ± 30 y (NIST 2015) — the currently best-measured value, and what you should use for a “true” age from first principles.

For classroom exercises use 5,700 y or 5,730 y. For anything you will compare to a published radiocarbon date, keep 5,568 y and report ages in “conventional radiocarbon years BP” — the calibration curves already account for the offset.

Common pitfalls

Zero-order half-life depends on [A]₀. Don’t quote $t_{1/2} = \ln 2 / k$ for a zero-order enzyme reaction.
Rate-constant units carry the order. A quoted $k$ with the wrong unit label is a strong signal the reaction is not first-order — check.
Fraction remaining vs. fraction reacted. “20 % of the original remains” is $f = 0.20$ ; “the reaction is 80 % complete” is the same physical situation. Confusing the two in time-to-fraction mode is the most common calculator misuse — this tool asks for fraction remaining.
Year conventions. This calculator uses the Julian year (1 yr = 365.25 d = 31,557,600 s exactly), which matches NIST SP 811 and the IAU 1976 convention adopted by NNDC for radiological half-lives. Do not confuse with the tropical (365.24219 d) or Gregorian (365.2425 d) year for these purposes.

Frequently asked questions

Why doesn't first-order half-life depend on the initial amount?: For a first-order process the rate is proportional to how much reactant (or radioactive nuclide) is present: r = k[A]. Solving the differential equation gives [A]_t = [A]_0 · exp(−k·t), so the ratio [A]_t/[A]_0 depends only on k·t — the [A]_0 cancels. The time for the ratio to fall to 1/2, t_½ = ln(2)/k, therefore involves only k. Zero- and second-order half-lives DO depend on [A]_0 (t_½ = [A]_0/(2k) for zero-order; t_½ = 1/(k[A]_0) for second-order), and that is one way to identify the order experimentally: if repeated measurements of t_½ change as [A]_0 changes, the reaction is not first-order.
What is the difference between activity and amount for radioactive decay?: The number of radioactive nuclei N is the *amount* of material; the activity A = −dN/dt = λ·N is the number of disintegrations per second (measured in becquerels, Bq, or curies, 1 Ci = 3.7 × 10¹⁰ Bq). Both decay with the same half-life because λ (the decay constant) is a constant of the nuclide. So A_t/A_0 = N_t/N_0 = exp(−λ·t) = (1/2)^(t/t_½). You can therefore work the arithmetic in whichever the problem gives you — a mass, a mole count, or an activity — and this calculator's math is identical.
How is radioactive decay related to first-order chemical kinetics?: Both are described by the same differential equation: the rate is proportional to the amount present. For chemistry that's r = k[A] with rate constant k in s⁻¹; for a radionuclide it is dN/dt = −λN with decay constant λ in s⁻¹. The two constants play identical mathematical roles, and t_½ = ln(2)/k = ln(2)/λ. The physical mechanism is different — quantum-mechanical tunneling out of a nuclear potential well vs. classical activation-energy barrier crossing — but the observable macroscopic law is the same.
When does the half-life formula not apply?: The clean t_½ = ln(2)/k relation is a first-order result. Zero-order and second-order integrated rate laws produce concentration-dependent half-lives (see the equation panel above). Complex mechanisms — reversible reactions near equilibrium, autocatalytic reactions, chain reactions, saturable enzyme kinetics — do not admit a single half-life at all: the apparent t_½ changes as the reaction progresses. If your reactant is not decaying exponentially, no single number of the form 'ln(2)/k' will describe it. For radioactive decay, the exponential law is exact (a first-order quantum-mechanical process); for chemistry, verify first-order behaviour by plotting ln[A] vs. t and checking that the line is straight.
Which value should I use for the carbon-14 half-life?: Radiocarbon dating uses two conventions. The 'Libby half-life' 5,568 y (Libby 1949) is embedded in the calibration curves and reported ¹⁴C ages by convention — you keep using it so historical results stay comparable. The 'Cambridge half-life' 5,730 ± 40 y (Godwin 1962; adopted widely in physics texts) is closer to the modern best measurement. NIST's evaluated 2015 value is 5,700 ± 30 y. For teaching and back-of-envelope work, use 5,700 or 5,730 y; for anything you'll compare to a published radiocarbon calibration, keep 5,568 y as the conventional half-life and report the age as 'conventional radiocarbon years BP'.

Sources

Atkins & de Paula, Physical Chemistry, 11th ed. (2018), §17B — Integrated rate laws
Silberberg, Chemistry: The Molecular Nature of Matter and Change, 8th ed. (2018), §16.4 — Integrated rate laws
IUPAC Gold Book — half-life (H02749)
NNDC (Brookhaven National Lab) — Nuclear Wallet Cards & evaluated half-lives
NIST Radionuclide Half-Life Measurements — evaluated ¹⁴C half-life 5,700 ± 30 y
NIST SP 811 (2008), Appendix B — Julian year (1 a = 365.25 d)
Godwin, H. (1962), Nature 195, 984 — Half-life of ¹⁴C revised to 5,730 ± 40 y

Last updated: 2026-09-18.

Half-Life Calculator — First-Order Kinetics & Decay