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Concept Version 9
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Calculations Involving Half-Life and Decay-Rates

The half-life of a radionuclide is the time taken for half the radionuclide's atoms to decay.

Learning Objective

  • Explain what is a half-life of a radionuclide.


Key Points

    • The half-life is related to the decay constant as follows: $t_{1/2} = ln2/\lambda$.
    • The relationship between the half-life and the decay constant shows that highly radioactive substances are quickly spent while those that radiate weakly endure longer.
    • Half-lives of known radionuclides vary widely, from more than 1019 years, such as for the very nearly stable nuclide 209 Bi, to 10-23 seconds for highly unstable ones.

Terms

  • half-life

    the time required for half of the nuclei in a sample of a specific isotope to undergo radioactive decay

  • radionuclide

    A radionuclide is an atom with an unstable nucleus, characterized by excess energy available to be imparted either to a newly created radiation particle within the nucleus or via internal conversion.


Example

    • A sample of 14C, whose half-life is 5730 years, has a decay rate of 14 disintegrations per minute (dpm) per gram of natural carbon. An artefact is found to have radioactivity of 4 dpm per gram of its present C. How old is the artefact? We have: $N = N_o e^{-t/\tau} \text{ where } N/N_o=4/14≈0.286 $, $\tau = t_{1/2}/ln2 \approx 8267 \text{ years, } t=−\tau lnN/N_o≈10360 \text{ years.}$

Full Text

The half-life of a radionuclide is the time taken for half of the radionuclide's atoms to decay. Taking $\lambda$ to be the decay rate (number of disintegrations per unit time) , and $\tau$ the average lifetime of an atom before it decays, we have:

$N(t) = N_{0}e^{-\lambda t} = N_{0}e^{-t/\tau}$

The half-life is related to the decay constant by substituting the condition $N=N_o /2$ and solving for $t = t_{1/2}$:

$t_{1/2} = ln2/\lambda = \tau ln2$

A half-life must not be thought of as the time required for exactly half of the atoms to decay. 

The following figure shows a simulation of many identical atoms undergoing radioactive decay. Note that after one half-life there are not exactly one-half of the atoms remaining; there are only approximately one-half left because of the random variation in the process. However, with more atoms (the boxes on the right), the overall decay is smoother and less random-looking than with fewer atoms (the boxes on the left), in accordance with the law of large numbers.

Radioactive decay simulation

A simulation of many identical atoms undergoing radioactive decay, starting with four atoms (left) and 400 atoms (right). The number at the top indicates how many half-lives have elapsed

The relationship between the half-life and the decay constant shows that highly radioactive substances are quickly spent while those that radiate weakly endure longer. Half-lives of known radionuclides vary widely, from more than 1019 years, such as for the very nearly stable nuclide 209 Bi, to 10−23 seconds for highly unstable ones.

The factor of ln(2) in the above equations results from the fact that the concept of "half-life" is merely a way of selecting a different base other than the natural base e for the lifetime expression. The time constant τ is the e-1-life, the time until only 1/e remains -- about 36.8 percent, rather than the 50 percent in the half-life of a radionuclide. Therefore, τ is longer than t1/2. The following equation can be shown to be valid:

$N(t) = N_{0}e^{-t/\tau } = N_{0}2^{-t/t_{1/2}}$

Since radioactive decay is exponential with a constant probability, each process could just as easily be described with a different constant time period that (for example) gave its 1/3-life (how long until only 1/3 is left), or its 1/10-life (how long until only 1/10 is left), and so on. Therefore, the choice of τ and t1/2 for marker-times is only for convenience and for the sake of uploading convention. These marker-times reflect a fundamental principle only in that they show that the same proportion of a given radioactive substance will decay over any time period you choose.

Mathematically, the nth life for the above situation would be found by the same process shown above -- by setting $N = N_{0}/n$ and substituting into the decay solution, to obtain:

$t_{1/n} = \frac{lnn}{\lambda } = \tau lnn$

Half-life

Part of a series of videos on physics problem-solving. The problems are taken from "The Joy of Physics. " This one deals with radioactive half-life. The viewer is urged to pause the video at the problem statement and work the problem before watching the rest of the video.

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