Radioactive Decay

Radioactive decay is the random and spontaneous decay of the nucleus of an atom to become more stable, resulting in the emission of alpha, beta or gamma radiation. 

When looking closely at a radioactive source, it is impossible to predict which nucleus will decay and when it will decay. This is because the process is completely random. It is also important to remember that no changes in the conditions surrounding a source will affect its rate of decay. 

The activity (A) of a radioactive source is the average number of nuclei that decay per unit of time. Notice two things from this definition:

• Average – as radioactive decay is random, the number of nuclei that decay per unit of time varies. We can however state the average number of decaying nuclei per unit of time.
• The definition states per unit of time. This unit of time can vary depending upon which isotope is being observed and how quickly they decay. 

A radioactive source that is described as highly radioactive has a high activity. This is measured in Becquerels $text{(Bq)}$ where $bold{1} text{ Bq}= bold{1}$ decaying nuclei per second. 

The equation used to calculate activity is as follows:

$A=dfrac{Delta N}{Delta t}$

• $A=$ the activity in becquerels $text{(Bq)}$
• $Delta N=$ the number of decayed nuclei 
• $Delta t=$ the change in time in seconds $text{(s)}$

Example: Over a period of one hour, a GM counter detects $25,000$ counts of radiation. What is the activity of the source? The background radiation has been measured at $1$ count per second.

[3 marks]

Substitute the given values into the activity equation:

$begin{aligned} bold{A} &= bold{dfrac{Delta N}{Delta t}} &= dfrac{textcolor{f43364}{25, 000}}{textcolor{10a6f3}{60 times 60}} &=  bold{6.9} text{ Bq} end{aligned}$

Subtract the background radiation:

$6.9 – textcolor{00d865}{1} = bold{5.9} textbf{ Bq}$

We can also use the decay constant $(lambda)$ to describe how radioactive a source is. The decay constant is the probability that a nuclei will decay per second. Decay constant is related to activity using the following equation:

$A= lambda N$

• $A=$ the activity in becquerels $text{(Bq)}$
• $lambda =$ the decay constant per second $text{(s}^{-1}text{)}$
• $N=$ the number of nuclei in a sample

Example: A source contains $6 times 10^{24}$ undecayed nuclei. If the decay constant is $2.1 times 10^{-6} text{ s}^{-1}$, calculate the activity of the source.

[2 marks]

Substitute the values into the equation linking activity and decay constant:

$begin{aligned} bold{A} &= bold{lambda N} &= textcolor{f95d27}{2.1 times 10^{-6}} times textcolor{ffad05}{6 times 10^{24}} &= bold{1.3 times 10^{19}} textbf{ Bq} end{aligned}$

Because of the nature of exponential decay, it can be difficult to predict how many nuclei would be remaining in a sample after a certain amount of time. A calculation can be completed to calculate this number:

$N=N_0e^{-lambda t}$

• $N=$ the number of undecayed nuclei 
• $N_0=$ the initial number of undecayed nuclei
• $lambda =$ the decay constant per second $text{(s}^{-1}text{)}$
• $t=$ the time in seconds $text{(s)}$

As the activity of a source and the count rate also decays exponentially, the equation may also be written as:

$A=A_0 e^{- lambda t}$

• $A=$ the activity at a given time (t) in becquerels $text{(Bq)}$
• $A_0$ the initial activity in becquerels $text{(Bq)}$
• $lambda=$ the decay constant per second $text{(s}^{-1}text{)}$
• $t=$ the time in seconds $text{(s)}$

$C=C_0 e^{-lambda t}$

• $C=$ the count rate at a given time (t) $(text{s}^{-1}text{, min}^{-1}text{ or hour}^{-1} text{ etc})$
• $C_0=$ the initial count rate $(text{s}^{-1}text{, min}^{-1}text{ or hour}^{-1} text{ etc})$
• $lambda =$ the decay constant per second $text{(s}^{-1}text{)}$
• $t=$ time in seconds $text{(s)}$

Example: A source contains $6 times 10^{24}$. If the decay constant of $2.1 times 10^{-6} text{ s}^{-1}$, calculate the activity of the source after $1$ year.

[4 marks]

Convert $1$ year to seconds:

$t= textcolor{ffad05}{1 text{ year}} = 1 times 365 times 24 times 60 times 60 = textcolor{f95d27}{bold{31 , 536 , 000} textbf{ s} }$

Calculate the initial activity:

$begin{aligned} A &= lambda N A_0 &= textcolor{00bfa8}{2.1 times 10^{-6}} times textcolor{d11149}{6 times 10^{24}}) &= textcolor{aa57ff}{bold{1.26 times 10^{19}}} textbf{ Bq} end{aligned}$

Substitute values into the equation for decay in activity:

$begin{aligned} bold{A} &= bold{A_0 e^{-lambda t}} &= textcolor{aa57ff}{1.26 times 10^{19}} times e^{-(textcolor{00bfa8}{2.1 times 10^{-6}} times textcolor{f95d27}{31 , 536 , 000}}) &= bold{2.25 times 10^{-10}} textbf{ Bq} end{aligned}$