Grade 12

Advanced

7 min

Lesson 9 of 12

Radioactivity and Exponential Decay

Not Started

Studying the random nature of nuclear decay and the mathematical laws governing half-life.

How can we determine the age of a 50,000-year-old mammoth bone or predict the exact potency of a cancer-treating isotope, when the decay of any single atom is a completely unpredictable, random event?

Learning Objectives

1.Compare alpha, beta, and gamma radiation in terms of mass, charge, and ionizing power
2.Solve complex problems using the exponential decay law $N = N_0 e^{-\lambda t}$ and half-life equations
3.Model and interpret the activity of radioactive sources using graphical analysis and semi-logarithmic plots

The Three Messengers of Decay

Radioactive decay occurs when an unstable nucleus loses energy by emitting radiation. We categorize this into three primary types: **Alpha ( $\alpha$ ), Beta ( $\beta$ ), and Gamma ( $\gamma$ ). Alpha particles** are helium nuclei ( $^4_2He$ ), making them heavy and highly charged ( $+2e$ ). This results in high ionizing power but very low penetration. Beta particles are high-speed electrons or positrons; they are much lighter, have a charge of $\pm 1e$ , and possess moderate penetration. Finally, Gamma radiation consists of high-energy photons. With no mass or charge, $\gamma$ -rays have the lowest ionizing power but can penetrate several centimeters of lead. The behavior of these emissions is governed by the conservation of mass number and atomic number.

Quick Check

Which type of radiation would be the most difficult to shield against using only a sheet of aluminum?

Think about which particle interacts the least with matter.

Answer

Gamma radiation, because it has no charge or mass and the highest penetration power.

The Mathematics of Spontaneous Change

Nuclear decay is stochastic (random), but for a large number of atoms, the rate of decay is proportional to the number of nuclei present. This is expressed by the differential equation:

\frac{dN}{dt} = -\lambda N

where

N

is the number of undecayed nuclei and

\lambda

is the decay constant (units of

s^{-1}

). Integrating this gives the Exponential Decay Law:

N(t) = N_0 e^{-\lambda t}

Here,

N_0

is the initial quantity and

e

is Euler's number. This equation tells us that the population of unstable nuclei decreases by a constant percentage per unit of time, regardless of the starting amount.

A sample contains $1.0 \times 10^{20}$ undecayed nuclei with a decay constant $\lambda = 0.05 \text{ s}^{-1}$ . How many nuclei remain after 10 seconds?

1. Identify variables: $N_0 = 1.0 \times 10^{20}$ , $\lambda = 0.05$ , $t = 10$ . 2. Set up the equation: $N = (1.0 \times 10^{20}) e^{-(0.05)(10)}$ . 3. Calculate the exponent: $e^{-0.5} \approx 0.6065$ . 4. Final result: $N \approx 6.07 \times 10^{19}$ nuclei.

Half-Life and Activity

The half-life (

t_{1/2}

) is the time required for half of the radioactive nuclei in a sample to decay. It is inversely proportional to the decay constant:

t_{1/2} = \frac{\ln(2)}{\lambda} \approx \frac{0.693}{\lambda}

We also measure Activity (

A

), which is the number of decays per second, measured in Becquerels (Bq). Since

A = \lambda N

, activity follows the same exponential trend:

A = A_0 e^{-\lambda t}

. In a graph of

N

t

, the curve is an asymptote. However, if we plot

\ln(N)

t

, we get a straight line with a slope of

-\lambda

, allowing for precise experimental determination of the decay constant.

A medical tracer has a half-life of 6 hours. If the initial activity is $800 \text{ MBq}$ , what is the activity after 24 hours?

1. Determine the number of half-lives ( $n$ ): $n = \frac{24}{6} = 4$ . 2. Use the shortcut for whole half-lives: $A = A_0 (\frac{1}{2})^n$ . 3. Calculate: $A = 800 \times (\frac{1}{2})^4 = 800 \times \frac{1}{16}$ . 4. Result: $A = 50 \text{ MBq}$ .

Quick Check

If a sample's decay constant $\lambda$ increases, what happens to its half-life?

Look at the relationship

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

Answer

The half-life decreases, as the atoms are decaying at a faster rate.

Advanced Graphical Analysis

In advanced physics, we often linearize data to find constants. By taking the natural log of the decay equation:

\ln(N) = \ln(N_0) - \lambda t

This fits the linear form

y = mx + c

. By plotting experimental data of

\ln(A)

\ln(N)

against time, the gradient of the resulting line is exactly

-\lambda

. This method is superior to reading values directly from a curve because it averages out random fluctuations in the count rate, providing a more accurate value for the half-life.

A student plots $\ln(A)$ against time (in minutes) and finds a linear best-fit line with a gradient of $-0.0231 \text{ min}^{-1}$ . Calculate the half-life.

1. Identify the gradient: $m = -\lambda$ , so $\lambda = 0.0231 \text{ min}^{-1}$ . 2. Use the half-life formula: $t_{1/2} = \frac{0.693}{\lambda}$ . 3. Calculate: $t_{1/2} = \frac{0.693}{0.0231}$ . 4. Result: $t_{1/2} = 30 \text{ minutes}$ .

Key Takeaways

1Alpha, Beta, and Gamma radiation represent a trade-off between ionizing power (damage) and penetration (range).
2The decay constant $\lambda$ represents the probability of a single nucleus decaying per unit time.
3The exponential law $N = N_0 e^{-\lambda t}$ applies to any large population of identical radioactive nuclei.
4Linearizing decay data using $\ln(N)$ vs $t$ allows for the precise calculation of $\lambda$ from the slope.

Quick Check

Multiple Choice

Which of the following correctly ranks the radiation types from highest to lowest ionizing power?

Multiple Choice

If the activity of a sample drops from $1000 \text{ Bq}$ to $125 \text{ Bq}$ in 15 days, what is the half-life?

True/False

The decay constant $\lambda$ of a radioactive isotope can be changed by increasing the temperature or pressure of the sample.

What's Next?

Review Tomorrow

In 24 hours, try to derive the relationship between the decay constant $\lambda$ and half-life $t_{1/2}$ starting from the exponential decay equation.

Practice Activity

Find a graph of radioactive decay in a textbook and try to calculate the activity at a point exactly 2.5 half-lives from the start using $A = A_0 e^{-\lambda t}$ .

Browse

Browse

Radioactivity and Exponential Decay

Learning Objectives

The Three Messengers of Decay

The Mathematics of Spontaneous Change

Half-Life and Activity

Advanced Graphical Analysis

Key Takeaways

Quick Check

What's Next?

Radioactivity and Exponential Decay

Learning Objectives

The Three Messengers of Decay

The Mathematics of Spontaneous Change

Half-Life and Activity

Advanced Graphical Analysis

Key Takeaways

Quick Check

What's Next?