Half-life and radioactive decay

In a nutshell

Radioactive decay is completely random. The amount of nuclei in a radioactive substance will slowly decrease as it decays and the time it takes for the substance to reach half the original number of nuclei is called its half-life. This lesson will discuss half-life and radioactive decay, how to calculate the activity of an isotope and how to calculate the half-life of a substance.

Equations

Description	Equation
Activity and number of nuclei relation	$A = \lambda N$
Exponential decay of nuclei	$N = N_0e^{-\lambda t}$
Exponential decay of activity	$A = A_0e^{-\lambda t}$
Half-life and decay constant relation	$\lambda t_{1/2} = \ln(2)$

Definitions

Keyword	Definition
Half-life	The time it takes for the amount of nuclei in a radioactive substance to decrease by half.
Activity	The rate of radioactive decay.
Isotope	The names given to a group of nuclei with the same number of protons but differing numbers of neutrons.

Variable definitions

Quantity name	symbol	derived unit	si unit
$half\ life$	$t_{1/2}$		$s$
$activity$	$A$	$Bq$	$s^{-1}$
$decay\ constant$	$\lambda$		$s^{-1}$
$number\ of\ nuclei$	$N$

Decay and half-life

On average, radioactive decay is completely random. There is no way to predict which nucleus in a sample will decay next and every nucleus has the same chance of decaying per unit time. The decay of the nuclei is also not affected by any external factors, nor does it depend on other nuclei.

The half-life $t_{1/2}$ of a substance is the time it takes for the number of nuclei in a particular sample to decrease by half. A sample decays exponentially. After one half-life, the amount of nuclei will be halved and after two, the amount of nuclei will be halved again. So, after two half-lives, the amount of nuclei will be a quarter of the original.

A substance with a very short half-life will decay quickly, releasing lots of radiation to begin with and quickly giving off less and less as the sample decays. A substance with a long half-life will release radiation over a longer period of time, during which the number of nuclei will decrease slowly.

Half-life from a graph

The graph below shows the activity of a sample decreasing with time.

Physics; Radioactivity; KS5 Year 12; Half-life and radioactive decay

A	Activity
B	Time

The half-life $t_{1/2}$ can be determined by finding the point on the $y$ -axis where activity has dropped to half of its original value $\dfrac{A_0}{2}$ . A line can be drawn to the curve, and then from the curve down to the $x$ -axis to determine at what time this value of activity occurs. This time is the half-life $t_{1/2}$ .

Activity

Activity $A$ is also called the rate of radioactive decay and is measured in Becquerels, given the symbol $Bq$ . $1\,Bq$ indicates that one nucleus decays every second. The activity of a source will exponentially decay over time, and the half-life also refers to the time it takes for the activity to halve.

Example

Carbon-14 has a half-life of $5700\ years$ . What percentage of carbon-14 will remain after $17100\ years$ ?

$17100\ years$ corresponds to three half-lives of carbon-14, indicating that the amount of carbon-14 will have been halved three times in that time.

$12.5\%$ of the original carbon-14 will remain after $17100\ years$ .

The decay constant

For a sample with a number of undecayed nuclei $N$ , the number of nuclei that are decaying $\Delta N$ is proportional to both $N$ and the time elapsed $\Delta t$ . This is the case because a sample with less nuclei will have a lower activity than one with more nuclei.

$\dfrac{\Delta N}{\Delta t} \propto -N$

This value $\dfrac{\Delta N}{\Delta t}$ is the activity of the source and can be written as a constant for a known isotope. This is known as the decay constant, given the symbol $\lambda$ . This equation can now be rewritten in terms of activity and decay constant, which has units $s^{-1}$ .

$A = \lambda N$

Example

There are $2.0\times 10^{16}$ undecayed nuclei in a source with a decay constant of $1.0 \times 10^{-9}\, s^{-1}$ . What is the activity of the source?

First, write down the known values:

$N = 2.0 \times 10^{16}$ 

$\lambda = 1.0\times 10^{-9}$ 

Next, write down the equation for activity:

$A = \lambda N$ 

Substitute in the values and calculate the final answer:

$A = (1.0\times 10^{-9}) \times (2.0\times10^{16})$

$A = 2.0\times 10^{7}\,Bq$

The sample has an activity of $\underline{2.0\times 10^{7}\,Bq}$ .

Exponential decay

The solution to the decay equation relating time and number of undecayed nuclei is given by the following equation.

$N = N_0e^{-\lambda t}$

where $N_0$ is the initial number of undecayed nuclei, $e$ is the natural logarithmic base (with a value of $2.718$ ) and $t$ is the time elapsed. Since activity is directly proportional to number of undecayed nuclei, $A$ can be rewritten in an exponential form as well.

$A = A_0e^{-\lambda t}$

where $A_0$ is the sample's initial activity.

Decay constant and half-life are linked together by these logarithmic equations. The first equation shows that one half-life has passed when $N$ is equal to $\dfrac{N_0}{2}$ , so $t$ must equal $t_{1/2}$ at that time. Substitute these values back into the equation.

$\dfrac{N_0}{2} = N_0e^{-\lambda t_{1/2}}$

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

This equation can then be rewritten without the negative power.

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

Finally, take the natural logarithm of both sides to arrive at the final equation.

$\lambda t_{1/2} = \ln(2)$

Example

What is the half-life of a sample with a decay constant of $1.0\times10^{-9}\,s^{-1}$ ?

First, write down the known values:

$\lambda = 1.0\times 10^{-9}$ 

Next, write down the equation and rearrange for half-life:

$\lambda t_{1/2}=\ln(2)$

$t_{1/2} = \dfrac{\ln(2)}{\lambda}$ 

Finally, substitute in the values and calculate the answer:

$t_{1/2} = \dfrac{\ln(2)}{(1.0 \times 10^{-9}) }$

$t_{1/2} = 22.0\ years$ 

The half-life of the sample is $\underline{22.0\ years}$ .