RADIOACTIVE DECAY
- The Random Nature of Radioactive Decay
- Radioactive decay is defined as:
The spontaneous disintegration of a nucleus to form a more stable nucleus, resulting in the emission of an alpha, beta or gamma particle - The random nature of radioactive decay can be demonstrated by observing the count rate of a Geiger-Muller (GM) tube
- When a GM tube is placed near a radioactive source, the counts are found to be irregular and cannot be predicted
- Each count represents a decay of an unstable nucleus
- These fluctuations in count rate on the GM tube provide evidence for the randomness of radioactive decay
Characteristics of Radioactive Decay
- Radioactive decay is both spontaneous and random
- A spontaneous process is defined as:
A process which cannot be influenced by environmental factors - This means radioactive decay cannot be affected by environmental factors such as:
- Temperature
- Pressure
- Chemical conditions
- A random process is defined as:
A process in which the exact time of decay of a nucleus cannot be predicted - Instead, the nucleus has a constant probability, ie. the same chance, of decaying in a given time
- Therefore, with large numbers of nuclei it is possible to statistically predict the behaviour of the entire group
Exam Tip
Make sure you can define what constitutes a radioactive decay, a random process and a spontaneous decay – these are all very common exam questions!
ACTIVITY & THE DECAY CONSTANT
- Since radioactive decay is spontaneous and random, it is useful to consider the average number of nuclei which are expected to decay per unit time
This is known as the average decay rate
- As a result, each radioactive element can be assigned a decay constant
- The decay constant $\lambda$ is defined as:
The probability that an individual nucleus will decay per unit of time - When a sample is highly radioactive, this means the number of decays per unit time is very high
This suggests it has a high level of activity
- Activity, or the number of decays per unit time can be calculated using:
$
\mathrm{A}=\frac{\Delta N}{\Delta t}=-\lambda \mathrm{N}
$
- Where:
- $A=$ activity of the sample $(\mathrm{Bq})$
- $\Delta \mathrm{N}=$ number of decayed nuclei
- $\Delta \mathrm{t}=$ time interval (s)
- $\lambda=$ decay constant $\left(\mathrm{s}^{-1}\right)$
- $\mathrm{N}=$ number of nuclei remaining in a sample
- The activity of a sample is measured in Becquerels $(\mathrm{Bq})$
- An activity of $1 \mathrm{~Bq}$ is equal to one decay per second, or $1 \mathrm{~s}^{-1}$
- This equation shows:
- The greater the decay constant, the greater the activity of the sample
- The activity depends on the number of undecayed nuclei remaining in the sample
- The minus sign indicates that the number of nuclei remaining decreases with time however, for calculations it can be omitted
Worked example: Decay constant
Americium-241 is an artificially produced radioactive element that emits a-particles.
A sample of americium-241 of mass $5.1 \mu \mathrm{g}$ is found to have an activity of $5.9 \times 10^5 \mathrm{~Bq}$.
a) Determine the number of nuclei in the sample of americium-241.
b) Determine the decay constant of americium-241.
Answer/Explanation
Part (a)
Step 1: Write down the known quantities
$
\text { Mass }=5.1 \mu \mathrm{g}=5.1 \times 10^{-6} \mathrm{~g}
$
Molecular mass of americium $=241$
Step 2: Write down the equation relating number of nuclei, mass and molecular mass
$
\text { Number of nuclei }=\frac{\text { mass } \times N_A}{\text { molecular mass }}
$
where $\mathrm{N}_{\mathrm{A}}$ is the Avogadro constant
Step 3:
Calculate the number of nuclei
$
\text { Number of nuclei }=\frac{\left(5.1 \times 10^{-6}\right) \times\left(6.02 \times 10^{23}\right)}{241}=1.27 \times 10^{16}
$
Part (b)
Step 1: $\quad$ Write the equation for activity
Activity, $A=\lambda N$
Step 2:
Rearrange for decay constant $\lambda$ and calculate the answer
$
\lambda=\frac{A}{N}=\frac{5.9 \times 10^5}{1.27 \times 10^{16}}=4.65 \times 10^{-11} \mathrm{~s}^{-1}
$
The Exponential Nature of Radioactive Decay
- In radioactive decay, the number of nuclei falls very rapidly, without ever reaching zero
- Such a model is known as exponential decay
- The graph of number of undecayed nuclei and time has a very distinctive shape
Equations for Radioactive Decay
- The number of undecayed nuclei $\mathrm{N}$ can be represented in exponential form by the equation:
$
\mathbf{N}=\mathbf{N o e}^{-\lambda t}
$
- Where:
- $\mathrm{N}_0=$ the initial number of undecayed nuclei (when $\mathrm{t}=0$ )
- $\lambda=$ decay constant $\left(s^{-1}\right)$
- $\mathrm{t}=$ time interval $(\mathrm{s})$
- The number of nuclei can be substituted for other quantities, for example, the activity $A$ is directly proportional to $\mathrm{N}$, so it can be represented in exponential form by the equation:
$
A=A^{-\lambda 2 t}
$
- The received count rate $C$ is related to the activity of the sample, hence it can also be represented in exponential form by the equation:
$
\mathrm{C}=\mathrm{Coe}^{-\lambda t}
$
- The exponential function $e$
- The symbol e represents the exponential constant
- It is approximately equal to $\mathrm{e}=2.718$
- On a calculator it is shown by the button $\mathrm{e}^{\mathrm{x}}$
- The inverse function of $\mathrm{e}^{\mathrm{x}}$ is $\ln (\mathrm{y})$, known as the natural logarithmic function
- This is because, if $e^x=y$, then $x=\ln (y)$
Worked example: Exponential decay
Strontium-90 decays with the emission of a $\beta$-particle to form Yttrium-90. The decay constant of Strontium-90 is 0.025 year $^{-1}$. Determine the activity $A$ of the sample after 5.0 years, expressing the answer as a fraction of the initial activity $\mathrm{A}_0$.
Answer/Explanation
Step 1:
Write out the known quantities
Decay constant, $\lambda=0.025$ year $^{-1}$
Time interval, $\mathrm{t}=\mathbf{5 . 0}$ years
Both quantities have the same unit, so there is no need for conversion
Step 2:
Write the equation for activity in exponential form
$
A=A_0 \mathrm{e}^{-\lambda t}
$
Step 3:
Rearrange the equation for the ratio between $\mathrm{A}$ and $\mathrm{A}_0$
$
\frac{A}{A_0}=e^{-\lambda t}
$
Step 4:
Calculate the ratio $\mathrm{A} / \mathrm{A}_0$
$
\frac{A}{A_0}=\mathrm{e}^{-(0.025 \times 5)}=0.88
$
Therefore, the activity of Strontium-90 decreases by a factor of 0.88 , or $12 \%$, after 5 years
HALF-LIFE
- Half life is defined as:
The time taken for the initial number of nuclei to reduce by half - This means when a time equal to the half-life has passed, the activity of the sample will also half
- This is because activity is proportional to the number of undecayed nuclei, $\mathrm{A} \propto \mathrm{N}$
Calculating Half-Life
- To find an expression for half-life, start with the equation for exponential decay:
$
\mathbf{N}=\mathbf{N o e}^{-\lambda t}
$
- Where:
- $\mathrm{N}=$ number of nuclei remaining in a sample
- $\mathrm{N}_0=$ the initial number of undecayed nuclei $($ when $\mathrm{t}=0)$
- $\lambda=$ decay constant $\left(s^{-1}\right)$
- $\mathrm{t}=$ time interval $(\mathrm{s})$
- When time $t$ is equal to the half-life $t_{1 / 2}$, the activity $\mathrm{N}$ of the sample will be half of its original value, so $\mathrm{N}=1 / 2 \mathrm{~N}_0$
$
\frac{1}{2} \mathrm{~N}_0=\mathrm{N}_0 \mathrm{e}^{-\lambda t_1 / 2}
$
- The formula can then be derived as follows:
Divide both sides by $N_0$ : $\quad \frac{1}{2}=e^{-\lambda t_1 / 2}$
Take the natural log of both sides: $\quad \ln \left(\frac{1}{2}\right)=-\lambda t_{1 / 2}$
Apply properties of logarithms: $\quad \lambda t_{1 / 2}=\ln (2)$
- Therefore, half-life $t_{k / 2}$ can be calculated using the equation:
$
t_{1 / 2}=\frac{\ln 2}{\lambda} \simeq \frac{0.693}{\lambda}
$
- This equation shows that half-life $t_{1 / 2}$ and the radioactive decay rate constant $\lambda$ are inversely proportional
- Therefore, the shorter the half-life, the larger the decay constant and the faster the decay
Worked example: Calculating half-life
Strontium-90 is a radioactive isotope with a half-life of 28.0 years. A sample of Strontium-90 has an activity of $6.4 \times 10^9 \mathrm{~Bq}$. Calculate the decay constant $\lambda$, in $\mathrm{s}^{-1}$, of Strontium- 90 .
Answer/Explanation
Step 1:
Convert the half-life into seconds
$
28 \text { years }=28 \times 365 \times 24 \times 60 \times 60=8.83 \times 10^8 \mathrm{~s}
$
Step 2:
Write the equation for half-life
$
t_{1 / 2}=\frac{\ln 2}{\lambda}
$
Step 3:
Rearrange for $\lambda$ and calculate
$
\lambda=\frac{\ln 2}{t_{1 / 2}}=\frac{\ln 2}{8.83 \times 10^8}=7.85 \times 10^{-10} \mathrm{~s}^{-1}
$