Radioactive decay is when an unstable nucleus decays in a random manner.

In theory, there is no end to the life of a radioactive substance (the time it takes before the activity reaches zero)therefore the quantity used for dealing with the life of radioactive substances is the half life. Half life is defined as the time taken for the activity of the sample to halve. The half life remains the same throughout the life of the sample. Each type of element has a distinct half life, $t_{\frac {1}{2}}$ . As the activity of the sample is proportional to the number of radioactive nuclides present the half time can also be calculated by the time taken for half of the radioactive nuclides in the sample to decay.

If you plot a graph of the natural logarithm of N (the number of radioactive nuclides) against time you will see that there is a linear relationship between the two. The steeper the gradient the more quickly the substance will decay which means it has a shorter half life. Half life is proportional to ${\frac {1}{\lambda }}$ # Equations

${\frac {\Delta N}{\Delta t}}=-\lambda N$ $N=N_{0}e^{-\lambda t}$ $t_{\frac {1}{2}}={\frac {ln2}{\lambda }}$ $A=\lambda N$ # Derivations of equations

Note - given that all equations above are provided in the AQA Formula booklet (which is provided in the exam), it is not necessary to memorise all of the equations, but it is a good idea to learn what the different symbols stand for.

## Deriving $N=N_{0}e^{-\lambda t}$ We can use this differential equation to derive an equation for N.

${\frac {dN}{dt}}=-\lambda N$ Separating and integrating:

$\int {\frac {1}{N}}dN=\int {-\lambda }dt$ $ln(N)=-\lambda t+c$ $N=N_{0}e^{-\lambda t}$ ## Deriving $t_{1/2}={\frac {ln2}{\lambda }}$ Half life is the time taken for half of the atoms that were originally present in the sample to have decayed.

$N_{(t)}=N_{0}e^{-\lambda t}$ If we replace $N_{0}$ with 1, $N_{t}$ with ${\frac {1}{2}}$ , and t with $t_{\frac {1}{2}}$ we get:

${\frac {1}{2}}=e^{-\lambda t_{\frac {1}{-2}}}$ $-\lambda t_{\frac {1}{2}}=ln{\frac {1}{2}}$ $t_{\frac {1}{2}}={\frac {ln2}{\lambda }}$ If you wanted to know the ${\frac {1}{n}}$ of a substance, you can replace the two with a 'n'

$t_{\frac {1}{n}}={\frac {ln(n)}{\lambda }}$ 