Problem 36 - Chemistry

Question

Radioactive copper-64 decays with a half-life of 12.8 days. a. What is the value of \(k\) in \(\mathrm{s}^{-1} ?\) b. A sample contains 28.0 \(\mathrm{mg}^{64} \mathrm{Cu}\) . How many decay events will be produced in the first second? Assume the atomic mass of \(^{64} \mathrm{Cu}\) is 64.0 \(\mathrm{u} .\) c. A chemist obtains a fresh sample of \(^{64} \mathrm{Cu}\) and measures its radioactivity. She then determines that to do an experiment, the radioactivity cannot fall below 25\(\%\) of the initial measured value. How long does she have to do the experiment?

Step-by-Step Solution

Answer

a. The value of \(k\) in \(\mathrm{s}^{-1}\) is \(\frac{\ln 2}{1105920 \, \text{s}}\). b. In the first second, \(-\frac{\ln 2}{1105920 \, \text{s}} \times 2.63 \times 10^{23} \, \text{atoms}\) decay events will occur. c. The chemist has \(\frac{\ln 0.25 \cdot 1105920 \, \text{s}}{\ln 2}\) seconds to complete the experiment while the radioactivity remains above 25% of the initial value.

Step 1: Find the decay constant.

To find the decay constant, we can use the half-life formula: \[t_\frac{1}{2} = \frac{\ln 2}{k}\], where \(t_\frac{1}{2}\) is the half-life. We know that the half-life of copper-64 is 12.8 days. Let's convert it to seconds by multiplying by 24 hours/day, 60 minutes/hour, and 60 seconds/minute: \[12.8 \, \text{days} = 12.8 \times 24 \times 60 \times 60 \, \text{s} = 1105920 \, \text{s}\]. Now we can solve for \(k\): \[k = \frac{\ln 2}{t_\frac{1}{2}} = \frac{\ln 2}{1105920 \, \text{s}}\].

Step 2: Calculate the number of decay events.

The given sample has a mass of 28.0 mg and an atomic mass of 64.0 u. To find the number of copper-64 atoms in the sample, we first need to convert the mass of the sample to moles, using the atomic mass as a conversion factor: \[\text{moles of }^{64}\mathrm{Cu} = \frac{28.0 \, \text{mg}}{64.0 \, \text{u/mol}} = 0.4375 \, \text{mol}\]. Next, we need to find the number of atoms in 0.4375 moles using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol): \[N_0 = 0.4375 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 2.63 \times 10^{23} \, \text{atoms}\]. Now we can find the number of decay events, which is equal to the rate of change of radioactive atoms with respect to time. First, find the derivative of the radioactive decay equation with respect to time: \[\frac{dN(t)}{dt} = -k \cdot N_0 \cdot e^{-k \cdot t}\]. At \(t = 0\), the rate of decay events is: \[\frac{dN(0)}{dt} = -k \cdot N_0 = -\frac{\ln 2}{1105920 \, \text{s}} \times 2.63 \times 10^{23} \, \text{atoms}\].

Step 3: Find the time for radioactivity to fall below a certain percentage.

To find the time at which the radioactivity falls below 25% of the initial radioactivity value, we can set \(N(t) = 0.25 \cdot N_0\) and solve for \(t\): \[0.25 \cdot N_0 = N_0 \cdot e^{-k \cdot t}\]. Taking the natural logarithm of both sides and simplifying: \[\ln 0.25 = -k \cdot t\]. Now, divide both sides by \(-k\) and substitute the value of \(k\) from Step 1: \[t = \frac{\ln 0.25}{-k} = \frac{\ln 0.25 \cdot 1105920 \, \text{s}}{\ln 2}\]. Now you can calculate each value to obtain the results: a. The value of \(k\) in \(\mathrm{s}^{-1}\) is \(\frac{\ln 2}{1105920 \, \text{s}}\). b. In the first second, \(-\frac{\ln 2}{1105920 \, \text{s}} \times 2.63 \times 10^{23} \, \text{atoms}\) decay events will occur. c. The radioactivity of the sample needs to remain above 25% of the initial value for \(\frac{\ln 0.25 \cdot 1105920 \, \text{s}}{\ln 2}\) seconds.

Key Concepts

Half-life

The concept of half-life is fundamental in understanding radioactive decay. Half-life refers to the time it takes for half of a radioactive substance to decay. For copper-64, this time is 12.8 days.

To put it simply, if you begin with 100 grams of copper-64, only 50 grams would remain after 12.8 days, as the other 50 grams would have decayed into another element or isotope. After another 12.8 days, you would have 25 grams remaining.

Half-life is significant because it allows us to predict how quickly or slowly a radioactive substance will decay.

• It's used in carbon dating to determine the age of ancient objects.
• It's critical in medical treatments that involve radioactive isotopes, helping to calculate the necessary dose.
• It plays a role in nuclear power, where understanding the decay of materials is essential for safety.

The formula used to calculate half-life is given by: \[t_{\frac{1}{2}} = \frac{\ln 2}{k}\] where \(t_{\frac{1}{2}}\) is the half-life, and \(k\) is the decay constant. This relationship shows a direct correlation between half-life and decay constant, with each helping to understand the decay process.

Decay Constant

The decay constant, denoted as \(k\), is a measure of the probability of decay per unit time. It helps express how fast a radioactive substance disintegrates.

In mathematical terms, the decay constant can be thought of as the component that quantifies the rate of radioactive decay. It is used in decay equations, often allowing us to predict how much of a substance remains at a given time, or how many decay events occur in a specific timeframe.

The relationship between half-life and decay constant is crucial; as defined by the formula: \[k = \frac{\ln 2}{t_{\frac{1}{2}}}\] where \(t_{\frac{1}{2}}\) is the half-life of the material. This formula makes it clear that a longer half-life corresponds to a smaller decay constant, indicating slower decay, and vice versa. Understanding \(k\) is vital in various fields:

• In nuclear medicine, for calculating the timing and dosages of treatments.
• In radioecology, for predicting the movement and impact of radioactive substances in the environment.
• In physics, to provide insights on the stability of isotopes.

Knowing \(k\) allows for accurately quantifying the behaviors of radioactive materials under different conditions.

Radioactivity

Radioactivity is the process by which unstable atomic nuclei lose energy by emitting radiation. This process can result in various types of radiation: alpha particles, beta particles, or gamma rays.

Radioactive decay is unpredictable for individual atoms, but predictable when considering large numbers of atoms. This predictability is why terms like decay constant and half-life are integral—these concepts help describe what happens as samples of radioactive material decay.

Radioactivity has numerous applications in modern science and technology:

• In medicine, for diagnosing and treating diseases such as cancer with techniques like PET scans and radiation therapy.
• In industry, where it helps inspect welds and metal integrity in construction and manufacturing.
• In energy production, as it underpins nuclear reactors and power plants.
• In archaeology, where it's used for dating ancient artifacts through carbon dating.

It's crucial to handle radioactive materials with care due to their potential health risks. Safety protocols are established to limit exposure and ensure that the benefits of radioactivity can be harnessed without undue harm to people or the environment.