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Half-life

From Biocrawler, the free encyclopedia.

For other uses, see Half-life (disambiguation).

The half-life of a radioactive substance is the time required for half of a sample to undergo radioactive decay.

More generally, for a quantity subject to exponential decay, the half-life is the time required for the quantity to fall to half of its initial value. (This article is a narrow discussion of half-life. For phenomena where half-life is applied, see "Related topics" below.)

After # of Half-lives	Percent of quantity remaining
0	100%
1	50
2	25
3	12.5
4	6.25
5	3.125
6	1.5625
7	0.78125%

The table at right shows the reduction of the quantity in terms of the number of half-lives elapsed.

Quantities subject to exponential decay are commonly denoted by the symbol N. (This convention suggests a decaying number of discrete items. This interpretation is valid in many, but not all, cases of exponential decay.) If the quantity is denoted by the symbol N, the value of N at a time t is given by the formula:

$N(t) = N_0 e^{-\lambda t} \,$

where

$N 0$ is the initial value of N (at t=0)
λ is a positive constant (the decay constant).

When t=0, the exponential is equal to 1, and N(t) is equal to $N 0$ . As t approaches infinity, the exponential approaches zero.

In particular, there is a time $t_{1/2} \,$ such that:

$N(t_{1/2}) = N_0\cdot\frac{1}{2}$

Substituting into the formula above, we have:

$N_0\cdot\frac{1}{2} = N_0 e^{-\lambda t_{1/2}} \,$

$e^{-\lambda t_{1/2}} = \frac{1}{2} \,$

$- \lambda t_{1/2} = \ln \frac{1}{2} = - \ln{2} \,$

$t_{1/2} = \frac{\ln 2}{\lambda} \,$

Thus the half-life is 69.3% of the mean lifetime.

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Decay by two or more processes

A radioactive element may decay via two or more different processes. These processes may have different probabilities of occuring, and thus there is also a different half-life associated with each process.

As an example, for two decay modes, the ammount of substance left after time t is given by

$N(t) = N_0 e^{-\lambda _1 t} e^{-\lambda _2 t} = N_0 e^{-(\lambda _1 + \lambda _2) t}$