Question

The specific activity of a radioactive material is the number of disintegrations per second per gram of radioactive atoms. a) Given the half-life of \({ }_{6}^{14} \mathrm{C}\) of \(5730 \mathrm{yr},\) calculate the specific activity of \({ }_{6}^{14} \mathrm{C}\). Express your result in disintegrations per second per gram, becquerel per gram, and curie per gram. b) Calculate the initial activity of a 5.00 -g piece of wood. c) How many \({ }_{6}^{14} \mathrm{C}\) disintegrations have occurred in a \(5.00-\mathrm{g}\) piece of wood that was cut from a tree on January \(1,1700 ?\)

Short answer

Question: Calculate the number of ${}_{6}^{14}\mathrm{C}$ disintegrations that occurred in a 5.00-g piece of wood since January 1, 1700. Answer: To find the number of ${}_{6}^{14}\mathrm{C}$ disintegrations since January 1, 1700, we need to follow these steps: 1. Calculate the decay constant from the half-life of ${}_{6}^{14}\mathrm{C}$. 2. Calculate the specific activity in disintegrations per second per gram. 3. Convert the specific activity to becquerel per gram and curie per gram. 4. Calculate the initial activity of a 5.00-g piece of wood. 5. Calculate the number of ${}_{6}^{14}\mathrm{C}$ disintegrations that occurred in a 5.00-g piece of wood since January 1, 1700. Using these steps, we can determine the number of disintegrations as \(N_\text{disintegrations} = (A_\text{wood} - A_\text{remaining}) \times t_\text{elapsed}\).

Step-by-step solution

1. Calculate the decay constant

The decay constant (λ) is related to the half-life (T) by the formula: \(λ = \frac{ln(2)}{T}\) Given the half-life of \({}_{6}^{14}\mathrm{C}\) of 5730 years, we can calculate the decay constant as: \(λ = \frac{ln(2)}{5730 \mathrm{yr}}\)

2. Calculate the specific activity

The specific activity (A) is given by the formula: \(A = λN\) where N is the number of radioactive atoms per gram. The nuclear mass of \({}_{6}^{14}\mathrm{C}\) is approximately 14 grams per mole. Since there are Avogadro's number (\(6.022 \times 10^{23}\)) of atoms in a mole, we can calculate N as: \(N = \frac{6.022 \times 10^{23} \text{ atoms}}{14 \text{ grams}}\) Using the calculated decay constant and number of atoms per gram, the specific activity of \({}_{6}^{14}\mathrm{C}\) is: \(A = λN\)

3. Convert the specific activity to becquerel and curie

The specific activity is given in disintegrations per second per gram, but we also need to express it in becquerel per gram (Bq/g) and curie per gram (Ci/g). 1 Bq = 1 disintegration per second, so the specific activity in Bq/g is the same as in disintegrations per second per gram. To convert to curies, use the relation: 1 Ci = 3.7 × 10¹⁰ Bq, so: \(\frac{A (\text{Ci/g})}{A (\text{Bq/g})} = \frac{3.7 \times 10^{10} \text{ Bq}}{1 \text{ Ci}}\)

4. Calculate the initial activity of a 5.00-g piece of wood

The activity of a 5.00-g piece of wood is given by the product of the specific activity and the mass: \(A_\text{wood} = A \times m\) where \(m = 5.00 \, \text{g}\).

5. Calculate the number of disintegrations since January 1, 1700

The elapsed time since January 1, 1700, is: \(t_\text{elapsed} = 2022-1700 = 322 \;\text{years}\) The remaining activity after the elapsed time can be calculated using the formula: \(A_\text{remaining} = A_\text{wood}e^{-λt_\text{elapsed}}\) The number of disintegrations is equal to the difference between the initial activity and the remaining activity multiplied by the elapsed time: \(N_\text{disintegrations} = (A_\text{wood} - A_\text{remaining}) \times t_\text{elapsed}\)

Key concepts

Specific Activity

The concept of specific activity is vital in understanding radioactive decay calculations. Specifically, it refers to the rate at which disintegrations occur in a given quantity of radioactive substance. To simplify, imagine you have a gram of the substance; specific activity tells you how many atoms in that gram are falling apart and becoming something else every second.

Let's bring this to life with a common example: carbon-14 (\({}_{6}^{14}\mathrm{C}\)). Scientists use its specific activity to date ancient artifacts, a technique known as radiocarbon dating. This simple yet crucial measure helps us to not just date ancient materials but also understand the dynamics of radioactive materials in medical applications, such as in cancer treatments.

Half-Life

Half-life is a term that sounds as it implies - it's the amount of time it takes for one half of a radioactive sample to decay. It's like a clock that doesn't tell time in hours or minutes but in the life of atoms. For \({}_{6}^{14}\mathrm{C}\), that time is 5730 years.

Why is this important? Well, it allows us to predict how a radioactive material will behave over time. By knowing the half-life, scientists can calculate when a sample will decay to a point where it's no longer detectable or when it's safe to handle after being used in medical treatments.

Radioactive Disintegrations

Radioactive disintegrations are the very heart of what it means to be radioactive. At the atomic level, disintegration is the breakdown of an unstable atomic nucleus into a more stable one. This breakdown releases energy and often a particle or two, like in carbon-14, which decays by emitting beta particles.

In practical terms, counting these disintegrations is how we measure radioactivity. The exercise you're working on deals with calculating how many disintegrations have occurred over time, which not only applies to dating artifacts but is also the basis for understanding radiation doses in medical and industrial applications.

Decay Constant

Decay constant, represented by the Greek letter lambda (\(\lambda\)), is the probability per unit time that an atom will decay. Think of it as a statistical average for a very large number of radioactive atoms. If half-life is the stopwatch, decay constant is the pace at which that stopwatch ticks.

Calculating the decay constant is a critical step because, from there, you're able to figure out the specific activity of the substance. It's like figuring out the heartbeat of the radioactivity, which helps you predict how it will behave over time, and therefore how to safely and effectively manage the use of this substance in various fields.

Avogadro's Number

Avogadro's number might sound a bit out of place in a discussion about radioactivity, but it's actually a linchpin in these calculations. It’s the link that turns abstract concepts into something measurable. Avogadro's number, which is approximately 6.022 x 10^23, tells us how many particles are in one mole of a substance.

In the context of our \({}_{6}^{14}\mathrm{C}\) example, this colossal number helps us determine the number of carbon-14 atoms in a gram of material. This tells us exactly how 'rich' our sample is with radioactive atoms, enabling us to calculate its specific activity and from there, its history and behavior over time.