Radioactive Decay and Case Studies - lect 4

Radioactive Decay Calculations

Equation for Radioactive Decay: N = N_0 e^{-kt}
- N: Amount at time t
- N_0: Initial amount
- k: Decay constant
- t: Time
Rearranging the Decay Equation:
- A document on Moodle explains how to rearrange the decay equation.
- Alternative approach: Learn the different versions of the equation.
Scenario 1:
- Initial counts (N0): 25
- Counts after 10 days (N): 107
- Time (t): 10 days
- Variable to find: k (decay constant), then calculate half-life in days
Calculating the Decay Constant (k):
- k = -\frac{\ln(\frac{N}{N_0})}{t}
- k = -\frac{\ln(\frac{107}{25})}{10} = 0.145
Calculating Half-Life:
- t_{1/2} = \frac{\ln(2)}{k} = \frac{0.693}{k}
- t_{1/2} = \frac{0.693}{0.0155} = 44.6 days
Unit Consistency:
- Units must be consistent throughout the calculation.
- Days for time and counts per second (or mass) for amount are okay as long as they are consistent.

Given:
- Half-life of Iodine-131: 8.04 days
- Target: Find the time it takes for the activity to fall to 1% of its initial value.
Step 1: Calculate the Rate Constant (k)
- k = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{8.04} = 0.08619 \text{ days}^{-1}
Step 2: Use the Time Equation
- t = -\frac{\ln(\frac{N}{N_0})}{k}
- The ratio \frac{N}{N_0} = 0.01 (1% of initial value)
Step 3: Calculate Time (t)
- t = -\frac{\ln(0.01)}{0.08619} = 53.4 days

Problem: What percentage of carbon-14 (half-life: 5730 years) remains after one year?
Goal: Find the ratio \frac{N}{N_0} after 1 year.
Equation:
- N = N_0 e^{-kt}
- \frac{N}{N_0} = e^{-kt}
Step 1: Calculate the Rate Constant (k)
- k = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{5730} = 1.21 \times 10^{-4} \text{ years}^{-1}
Step 2: Calculate the Ratio
- \frac{N}{N_0} = e^{-(1.21 \times 10^{-4})(1)}
- \frac{N}{N_0} = 0.99988
Step 3: Convert to Percentage
- 0.99988 \times 100 = 99.988\%
Sanity Check:
- After one half-life (5730 years), the ratio would be 0.5 or 50%.
- After one year, expect the value to be very close to 100% since very little decay occurs.

Lindow Moss: A peat harvesting site near Manchester.
Discovery in 1983: Workers found a head and reported it to the police.
Peter Ray Bart Case:
- In the 1950s, Peter Ray Bart married Malaika, who disappeared shortly after.
- Police suspected him but lacked evidence.
- Homosexuality was illegal at the time, and the marriage was a cover.
- He confessed that Malaika threatened to expose him, leading to her murder.
The Head Discovery:
- The head was discovered in 1983, and police questioned Peter Ray Bart.
- He confessed, believing it was his wife's head, due to its seemingly fresh appearance.
- Carbon dating revealed the head was 1750 years old.
Peat Preservation:
- Peat environments are known to preserve bodies due to unique decomposition conditions.
Legal Outcome:
- Peter Ray Bart tried to retract his confession, but the judge convicted him of murder.

Discovery: Found in Lindow Moss during peat cutting operations.
Preservation: The body is well-preserved; toenails, hair follicles, and stubble are visible.
Sphagnum Moss:
- Sphagnum moss in bogs produces chemicals that convert skin to leather, aiding preservation.
Cause of Death:
- Lindow Man suffered multiple injuries: struck on the head, garroted, and neck broken.
Ritual Killing:
- Eamon Kelly suggests these were ritual sacrifices to the gods.
- Bogs on tribal boundaries were seen as access points to the other world.
Lindau Woman:
- Another body found, highlighting the significance of carbon dating for accurate dating.

Vorum Perce Medieval Village: A pit of mutilated bodies was discovered.
Objective: Determine when the people in the pit died to understand potential events.
Radiocarbon Dating:
- Organic remains like human bone can be dated using radiocarbon dating.
- Living organisms absorb carbon-14, which decays at a known rate after death.
- By measuring the remaining carbon-14, the time of death can be estimated.
Samples Taken:
- Samples were taken from bones with cut marks, broken bones, and burnt bones.
Results:
- Radiocarbon dating revealed a date range spanning two to three centuries.
- This indicated that the bodies were not killed and dismembered simultaneously, suggesting repeated patterns of violence over centuries.

Problem: A 1-gram sample of wood undergoes 7900 carbon-14 disintegrations in 20 hours. A 1-gram sample of modern carbon undergoes 18,400 disintegrations in the same period. Calculate the age of the wood.
Key Information:
- Sample mass (1 gram) and measurement period (20 hours) are consistent and can be treated as ratios.
Estimating the Answer:
- Modern carbon (N0): 18,400 disintegrations
- Old wood (N): 7,900 disintegrations
- After one half-life, modern carbon would have 9,200 disintegrations. Since the wood is below this, it's more than one half-life old (5730 years).
Calculations:
- Use the equation: t = -\frac{\ln(\frac{N}{N_0})}{k}
- First, find k: k = \frac{\ln(2)}{5730} = 1.21 \times 10^{-4}
- Then, calculate t: t = -\frac{\ln(\frac{7900}{18400})}{1.21 \times 10^{-4}} = 6941 years.

Question 1: A sample contains 120 grams of a radioactive isotope. After 15 years, only 30 grams of the isotope remain. What is the half-life of the isotope?
Question 2: The half-life of a certain radioactive substance is 3.6 hours. If you start with 500 grams of the substance, how much will remain after 12 hours?
Question 3: A rock sample is found to contain 65% of its original potassium-40. The half-life of potassium-40 is 1.25 \times 10^9 years. How old is the rock?
Question 4: A bone fragment found in an archaeological dig is carbon-dated. The carbon-14 content is found to be 22% of that in living bone. How old is the bone fragment? (Half-life of carbon-14 is 5730 years)
Question 5: A radioactive tracer has a half-life of 2.0 days. How long will it take for 3/4 of the initial amount to decay?
Question 6: A 200g sample of a radioactive isotope decays to 50g in 20 minutes. What is the half-life of this isotope?
Question 7: How many years will it take for 100 grams of Plutonium-239 (half-life 24,100 years) to decay to 20 grams?
Question 8: A piece of ancient wood contains 15% of its original Carbon-14. How old is the wood?
Question 9: If a radioactive substance has a half-life of 72 hours, what fraction of the initial sample will remain after 18 hours?
Question 10: A medical isotope has a half-life of 6 hours. If a patient is injected with 40 mCi, how much will remain after 24 hours?