# Half-Life ## Understanding Half-Life

The time taken for the initial number of nuclei to reduce by half

- When a time equal to the half-life has passed, the activity of the sample will be half of its original value - This is because activity is proportional to the number of undecayed nuclei: $A \propto N$ ### Activity Changes Over Time - After one half-life passes, activity falls to half its original value - After two half-lives pass, activity falls to a quarter of its original value - After three half-lives pass, activity falls to an eighth of its original value - The pattern continues exponentially | **Time elapsed** | **Fraction remaining** | **Percentage remaining** | |------------------|------------------------|--------------------------| | 0 half-lives | $1$ | 100% | | 1 half-life | $\frac{1}{2}$ | 50% | | 2 half-lives | $\frac{1}{4}$ | 25% | | 3 half-lives | $\frac{1}{8}$ | 12.5% | | 4 half-lives | $\frac{1}{16}$ | 6.25% | ### Visualizing Exponential Decay The chart below shows how activity decreases exponentially with time. Notice how the activity halves at regular intervals (each half-life):

## Mathematical Expression of Half-Life ### Starting with the Exponential Decay Equation - The number of undecayed nuclei can be represented by: $$N = N_0 e^{-\lambda t}$$ - Where: - $N$ = number of nuclei remaining in a sample - $N_0$ = initial number of undecayed nuclei (when $t = 0$) - $\lambda$ = decay constant ($\text{s}^{-1}$) - $t$ = time interval (s) ### Deriving the Half-Life Formula - When time $t$ equals the half-life $t_{1/2}$, the number of nuclei $N$ will be half the original value - So, $N = \frac{1}{2}N_0$ **Step 1:** Substitute into the exponential decay equation $$\frac{1}{2}N_0 = N_0 e^{-\lambda t_{1/2}}$$ **Step 2:** Divide both sides by $N_0$ $$\frac{1}{2} = e^{-\lambda t_{1/2}}$$ **Step 3:** Take the natural logarithm of both sides $$\ln\left(\frac{1}{2}\right) = -\lambda t_{1/2}$$ **Step 4:** Apply properties of logarithms $$\lambda t_{1/2} = \ln(2)$$ **Step 5:** Rearrange for half-life

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

### Relationship Between Half-Life and Decay Constant - Half-life $t_{1/2}$ and decay constant $\lambda$ are **inversely proportional** - The shorter the half-life, the larger the decay constant - The longer the half-life, the smaller the decay constant - This means: - Short half-life → rapid decay → high decay rate - Long half-life → slow decay → low decay rate ## Activity and Half-Life - Activity can also be expressed using the exponential form: $$A = A_0 e^{-\lambda t}$$ - Where: - $A$ = activity at a certain time $t$ (Bq) - $A_0$ = initial activity (Bq) - $\lambda$ = decay constant ($\text{s}^{-1}$) - $t$ = time interval (s) - Since activity is proportional to the number of undecayed nuclei, it follows the same exponential decay pattern ## Worked Example: Calculating Activity After a Given Time **Question:** Strontium-90 is a radioactive isotope with a half-life of 28.0 years. A sample of strontium-90 has an initial activity of $A_0$. Calculate the activity $A$ of the sample after 5.0 years, expressing the answer as a fraction of the initial activity. **Solution:** **Step 1:** Convert the half-life into seconds - $t_{1/2} = 28 \text{ years}$ - $t_{1/2} = 28 \times (365 \times 24 \times 60 \times 60)$ - $t_{1/2} = 8.83 \times 10^8 \text{ s}$ **Step 2:** Calculate the decay constant $$\lambda = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{8.83 \times 10^8} = 7.85 \times 10^{-10} \text{ s}^{-1}$$ **Step 3:** Convert time interval to seconds - $t = 5.0 \text{ years} = 5.0 \times (365 \times 24 \times 60 \times 60)$ - $t = 1.58 \times 10^8 \text{ s}$ **Step 4:** Write the exponential activity equation $$A = A_0 e^{-\lambda t}$$ **Step 5:** Rearrange for the ratio $A/A_0$ $$\frac{A}{A_0} = e^{-\lambda t}$$ **Step 6:** Calculate the ratio $$\frac{A}{A_0} = e^{-(7.85 \times 10^{-10}) \times (1.58 \times 10^8)} = e^{-0.124} = 0.88$$ - Therefore, the activity of strontium-90 decreases to 88% of its original value (or decreases by 12%) after 5 years ## Practical: Determining the Half-Life of an Isotope ### Aim - To determine the half-life of an isotope such as protactinium-234 using experimental measurements ### Variables - **Independent variable:** Time, $t$ (s) - **Dependent variable:** Corrected count rate, $C$ - **Control variables:** - Radioactive source - Distance of GM tube to source - Location / background radiation ### Equipment | **Equipment** | **Purpose** | |---------------|-------------| | Sealed bottle containing uranium salt solution and protactinium-234 | To use as a source of radioactive emission | | Geiger-Muller tube and counter (or datalogger) | To measure the count rate of the radioactive source | | Stopwatch | To measure the same time interval for each count rate reading | | Tray lined with paper towel | To avoid contamination should the bottle leak | ### Method **Background information:** - In the lower layer of the sealed bottle, uranium-238 decays to thorium-234 and then to protactinium-234 - Shaking the bottle separates the isotopes as protactinium dissolves in the organic solvent, whereas thorium does not **Procedure:** 1. Connect the Geiger-Muller tube to the counter 2. Without any sources present, measure background radiation over a period of one minute 3. Repeat this three times and take an average 4. Subtract this background value from all subsequent readings 5. Shake the bottle gently for 15 seconds to dissolve the protactinium in the organic layer 6. Wait for the organic layer to float to the top 7. Align the GM tube with the organic layer and immediately start a stopwatch 8. Record the number of counts in 10 seconds 9. Repeat every 30 seconds for several minutes **Recording results:** | **Time, $t$ / s** | **Count rate / counts s$^{-1}$** | **Corrected count rate, $C$ / counts s$^{-1}$** | |-------------------|----------------------------------|------------------------------------------------| | 0 | | | | 30 | | | | 60 | | | | 90 | | | | ... | | | ### Analysis of Results

Always subtract background radiation from count rate measurements before analysis

- The count rate of the source decreases exponentially according to: $$C = C_0 e^{-\lambda t}$$ **Taking natural logarithms of both sides:** $$\ln(C) = \ln(C_0) - \lambda t$$ **Comparing to the equation of a straight line** ($y = mx + c$): - y-axis variable: $y = \ln(C)$ - x-axis variable: $x = t$ - Gradient: $m = -\lambda$ - y-intercept: $c = \ln(C_0)$ **Steps for analysis:** 1. Subtract the background radiation from each count rate reading to give the corrected count rate $C$ 2. Calculate $\ln(C)$ for each measurement 3. Plot a graph of $\ln(C)$ against time $t$ 4. Draw a best-fit straight line through the points 5. Calculate the gradient of the line 6. Determine the half-life using: $$t_{1/2} = -\frac{\ln(2)}{\text{gradient}} = \frac{0.693}{|\text{gradient}|}$$ - Since gradient $= -\lambda$, then $\lambda = -\text{gradient}$ ### Expected Graph The graph below shows the expected result when plotting $\ln(C)$ against time. The straight line with negative gradient confirms exponential decay:

### Evaluating the Experiment **Systematic Errors:** - Position the GM tube close to the bottle, but ensure they don't touch - This ensures only beta particles from protactinium are detected, not alpha particles from uranium (which are absorbed by the bottle) **Random Errors:** - Radioactive decay is random, so repeat readings are vital - The half-life of protactinium-234 is relatively short (about 1 minute), so more frequent readings of count rate may improve accuracy - Take multiple readings at each time interval and calculate an average **Safety Considerations:** - Protactinium-234 is relatively safe to use due to its short half-life - Ensure the bottle remains sealed at all times to prevent leakage - Handle the bottle with long tongs to maintain distance - Wear appropriate safety clothing: lab coat, gloves, and goggles - Store sources in a lead-lined container when not in use - Minimize exposure time to the radioactive source