Radioactivity is a natural process where unstable atomic nuclei spontaneously transform into more stable forms by emitting particles or radiation. This transformation is called radioactive decay. It happens because certain isotopes-versions of elements with different numbers of neutrons-are inherently unstable. Unlike many physical processes that can be predicted exactly, radioactive decay is inherently probabilistic. We cannot know exactly when a particular nucleus will decay, only the probability that it will decay over a certain time interval.
Two key concepts help us understand and quantify radioactive decay: half-life and activity. Half-life tells us how long it takes for half of a sample to decay, while activity measures how many decays occur per second. These ideas are foundational in nuclear physics and critical for applications such as medical imaging, radiocarbon dating, nuclear energy, and radiation safety.
Imagine you have a large number of identical radioactive nuclei, say \( N_0 \) at time \( t = 0 \). Because each nucleus has a fixed probability of decaying per unit time, the number of undecayed nuclei \( N \) decreases as time passes. The rate at which nuclei decay is proportional to how many remain. Mathematically, this is expressed by the differential equation:
\[ \frac{dN}{dt} = -\lambda N \]
Here, \( \lambda \) is called the decay constant and has units of inverse time (\( s^{-1} \), for example). It quantifies the probability of decay per nucleus per second. The negative sign indicates that \( N \) decreases over time.
Solving this equation (a standard first-order differential equation) gives us the number of undecayed nuclei at any time \( t \):
\[ N = N_0 e^{-\lambda t} \]
This is known as the exponential decay law. It tells us that the number of nuclei reduces exponentially over time, never dropping abruptly but slowing as fewer nuclei remain.
The half-life of a radioactive isotope, symbolized as \( T_{1/2} \), is defined as the time taken for half of the initial number of nuclei to decay. In other words, after a time \( T_{1/2} \):
\[ N = \frac{N_0}{2} \]
Using the exponential decay law, we substitute and find \( T_{1/2} \):
\[ \frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \]
Dividing both sides by \( N_0 \):
\[ \frac{1}{2} = e^{-\lambda T_{1/2}} \]
Taking the natural logarithm of both sides:
\[ \ln \frac{1}{2} = -\lambda T_{1/2} \Rightarrow -\ln 2 = -\lambda T_{1/2} \]
Thus, the formula for half-life is:
\[ T_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda} \]
This relationship shows that the half-life depends only on the decay constant and is independent of the initial amount or the physical or chemical state of the substance. It's a unique property of each radioactive isotope.
Activity (\( A \)) of a radioactive sample is defined as the number of decays occurring in the sample per unit time, typically per second. Since each decay corresponds to a nucleus disintegrating, the activity is related to how many nuclei are left and their decay probability.
The rate of change of nuclei is:
\[ \frac{dN}{dt} = -\lambda N \]
Since activity counts decays (positive quantity),
\[ A = -\frac{dN}{dt} = \lambda N \]
Activity decreases over time as nuclei decay, which follows the exponential law:
\[ A = A_0 e^{-\lambda t} \]
where \( A_0 = \lambda N_0 \) is the initial activity right at \( t = 0 \).
Units of Activity:
Activity gives a practical measure of how "radioactive" a sample is at any time.
graph TD N[Number of undecayed nuclei (N)] L[Decay constant (λ)] A[Activity (A = λN)] T[Time (t)] N -->|decreases over time| N L --> A N --> A T -->|exponential decay| N T -->|exponential decay| A
Step 1: Recall the formula relating half-life and decay constant: \( T_{1/2} = \frac{\ln 2}{\lambda} \).
Step 2: Substitute the value of \( \lambda \):
\[ T_{1/2} = \frac{0.693}{2.31 \times 10^{-4}} = 2995.7 \text{ seconds} \]
Step 3: Convert seconds to minutes for easier understanding:
\[ 2995.7 \text{ s} \div 60 = 49.93 \text{ minutes} \]
Answer: The half-life is approximately 50 minutes.
Step 1: Calculate the decay constant using \( \lambda = \frac{\ln 2}{T_{1/2}} \):
\[ \lambda = \frac{0.693}{20 \times 3600} = \frac{0.693}{72000} = 9.625 \times 10^{-6} \text{ s}^{-1} \]
Step 2: Convert 60 hours to seconds for consistency:
\[ 60 \times 3600 = 216000 \text{ s} \]
Step 3: Use the decay law \( N = N_0 e^{-\lambda t} \):
\[ N = 5 \times 10^{12} \times e^{-9.625 \times 10^{-6} \times 216000} \]
Calculate the exponent:
\[ -9.625 \times 10^{-6} \times 216000 = -2.08 \]
\[ N = 5 \times 10^{12} \times e^{-2.08} = 5 \times 10^{12} \times 0.125 \]
\[ N = 6.25 \times 10^{11} \text{ nuclei} \]
Answer: Approximately \( 6.25 \times 10^{11} \) nuclei remain undecayed after 60 hours.
Step 1: Calculate decay constant:
\[ \lambda = \frac{0.693}{4 \times 3600} = \frac{0.693}{14400} = 4.813 \times 10^{-5} \text{ s}^{-1} \]
Step 2: Convert 10 hours into seconds:
\[ 10 \times 3600 = 36000 \text{ s} \]
Step 3: Use \( A = A_0 e^{-\lambda t} \):
\[ A = 5000 \times e^{-4.813 \times 10^{-5} \times 36000} \]
Calculate exponent:
\[ -4.813 \times 10^{-5} \times 36000 = -1.732 \]
\[ A = 5000 \times e^{-1.732} = 5000 \times 0.177 \approx 885 \text{ Bq} \]
Answer: Activity after 10 hours is approximately 885 Bq.
Step 1: Use the activity-time relation:
\[ A = A_0 e^{-\lambda t} \Rightarrow \frac{A}{A_0} = e^{-\lambda t} \]
Step 2: Substitute known values:
\[ \frac{2000}{8000} = e^{-\lambda \times (3 \times 3600)} \Rightarrow 0.25 = e^{-10800 \lambda} \]
Step 3: Take natural logarithm on both sides:
\[ \ln 0.25 = -10800 \lambda \Rightarrow -1.386 = -10800 \lambda \]
\[ \lambda = \frac{1.386}{10800} = 1.283 \times 10^{-4} \text{ s}^{-1} \]
Step 4: Calculate half-life:
\[ T_{1/2} = \frac{0.693}{1.283 \times 10^{-4}} = 5400 \text{ seconds} \]
Convert to hours:
\[ \frac{5400}{3600} = 1.5 \text{ hours} \]
Answer: Decay constant \( \lambda = 1.283 \times 10^{-4} \text{ s}^{-1} \) and half-life \( T_{1/2} = 1.5 \) hours.
Step 1: Calculate the decay constant:
\[ \lambda = \frac{0.693}{6 \times 3600} = \frac{0.693}{21600} = 3.21 \times 10^{-5} \text{ s}^{-1} \]
Step 2: Convert 12 hours to seconds:
\[ 12 \times 3600 = 43200 \text{ s} \]
Step 3: Use decay law for activity:
\[ A = A_0 e^{-\lambda t} = 500 \times 10^{6} \times e^{-3.21 \times 10^{-5} \times 43200} \text{ Bq} \]
Calculate exponent:
\[ -3.21 \times 10^{-5} \times 43200 = -1.386 \]
\[ A = 500 \times 10^{6} \times e^{-1.386} = 500 \times 10^{6} \times 0.25 = 125 \times 10^{6} \text{ Bq} = 125 \text{ MBq} \]
Answer: After 12 hours, the activity reduces to 125 MBq.
Summary of Key Formulas:
When to use: Avoid confusing starting amounts with decay time when solving problems.
When to use: When isolating decay constant or time in exponentials.
When to use: To maintain unit consistency in calculations.
When to use: To quickly compare activity at different times without recalculating full decay law.
When to use: On time-pressed entrance exam questions for rapid estimates.
Progress tracking is paywalled — subscribe to mark subtopics as understood and save your streak.
Go to practice →