Question

Radioactive copper- 64 decays with a half-life of 12.8 days. a. What is the value of \(k\) in \(s^{-1} ?\) b. A sample contains \(28.0 \space\mathrm{mg}\space^{64} \mathrm{Cu}\). 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?

Short answer

a. The decay constant, \(k = 6.27 \times 10^{-7}\ s^{-1}\). b. The number of decay events in the first second is \(1.65 \times 10^{16}\) events. c. The chemist has 51.3 days to complete the experiment before the radioactivity falls below 25% of the initial value.

Step-by-step solution

Find the decay constant k

To find the decay constant (k), we will use the half-life formula: \[t_{1/2} = \frac{0.693}{k}\] Where \(t_{1/2}\) is the half-life (12.8 days) and k is the decay constant. We are asked to find the value of k in \(s^{-1}\), so we must first convert 12.8 days to seconds: \[12.8 \text{ days} \cdot 24 \frac{\text{hours}}{\text{day}} \cdot 60 \frac{\text{min}}{\text{hour}} \cdot 60 \frac{\text{s}}{\text{min}} = 1105920 \text{ s}\] Now, we can find the decay constant: \[k = \frac{0.693}{t_{1/2}} = \frac{0.693}{1105920 \text{ s}} = 6.27 \times 10^{-7}\ s^{-1}\]

Calculate the number of decay events in the first second

We can use the decay equation to find the number of decay events in the first second: \[N(t) = N_0 e^{-kt}\] Where \(N(t)\) is the number of events at time t, \(N_0\) is the initial number of atoms, and k and t are as defined in Step 1. We are given the mass of the sample (28.0 mg), so we can find the number of initial atoms by dividing the mass by the atomic mass of Copper-64 and multiplying by Avogadro's number (\(6.022 \times 10^{23} \text{ atoms/mol}\)): \[N_0 = \frac{28.0 \text{mg} \cdot 10^{-3} \frac{\text{g}}{\text{mg}}}{64.0 \frac{\text{g}}{\text{mol}}} \cdot 6.022 \times 10^{23} \text{ atoms/mol} = 2.63 \times 10^{22} \text{ atoms}\] Now, we can find the number of decay events in the first second by differentiating the decay equation with respect to time and evaluating it at t = 0: \[\frac{dN(t)}{dt} = -k N_0 e^{-kt}\] At t = 0: \[\frac{dN(0)}{dt} = -k N_0 e^{0} = -k N_0\] \[\frac{dN(0)}{dt} = -(6.27 \times 10^{-7} \space s^{-1})(2.63 \times 10^{22} \text{ atoms}) = -1.65 \times 10^{16} \text{ atoms/s}\] Since we are looking for the number of decay events in the first second, we take the absolute value: \[\text{Decay events in the first second} = 1.65 \times 10^{16}\]

Calculate the time for radioactivity to fall below 25% of the initial value

We are given that the radioactivity cannot fall below 25% of the initial measured value. We can find the time for this to happen using the decay equation: \[N(t) = \frac{1}{4} N_0\] \[e^{-kt} = \frac{1}{4}\] Take the natural logarithm of both sides: \[-kt = \ln{\frac{1}{4}} = -2 \ln{2}\] Solve for t: \[t = \frac{-2 \ln{2}}{-k} = \frac{-2 \ln{2}}{6.27 \times 10^{-7} \space s^{-1}} = 4436330 \text{ s}\] To make this time easier to interpret, we can convert it to days: \[t = 4436330 \text{ s} \cdot \frac{1 \text{ min}}{60 \text{ s}} \cdot \frac{1 \text{ hour}}{60 \text{ min}} \cdot \frac{1 \text{ day}}{24 \text{ hour}} = 51.3 \text{ days}\] The chemist has 51.3 days to do the experiment before the radioactivity falls below 25% of the initial value.

Key concepts

Half-life

Understanding the concept of half-life is crucial in the field of nuclear chemistry and radioactive decay calculations. Simply put, the half-life of a radioactive substance is the time it takes for half of the substance's atoms to decay. This period is unique to each radioactive isotope and can range from fractions of a second to millions of years.

In practical terms, if you start with a certain amount of a radioactive isotope, after one half-life, you'll have half the original amount left. After another half-life has passed, you'll have half of that amount, or one quarter of the starting amount. This process continues, with the quantity of the substance decreasing by half with each half-life.

The half-life equation \(t_{1/2} = \frac{0.693}{k}\) links the decay constant (\(k\)) to the half-life (\(t_{1/2}\)). Through this relationship, if we know one, we can determine the other. In educational materials, recognizing the significance of half-life helps students connect abstract decay constants to more tangible time intervals.

Decay Constant

The decay constant, represented by the symbol \(k\), is a probability factor that quantifies the average rate at which a radioactive isotope decays. Mathematically, it's the proportion of atoms that decay per unit time, expressing how quickly or slowly the decay process occurs. Determining the decay constant is a key step in predicting the behavior of a radioactive sample over time.

The formula \(k = \frac{0.693}{t_{1/2}}\) shows that the decay constant is inversely proportional to the half-life of the isotope – a higher decay constant means a shorter half-life and vice versa. To find \(k\) in \(s^{-1}\), it's essential to express the half-life in seconds, since the constant is a rate per second. This relationship is the backbone of calculations such as determining the number of decay events in a given timeframe or predicting the stability of a sample over time.

Radioactive Decay Events

Radioactive decay events refer to the individual transformations that occur when an unstable atomic nucleus releases energy to become more stable. These events can include the emission of alpha particles, beta particles, or gamma rays. The rate at which these decay events occur is of significant interest in various fields, such as medicine, archaeology, and environmental science, among others.

To calculate the number of decay events in a specific time frame, we use the decay equation \(N(t) = N_0 e^{-kt}\). Here, \(N(t)\) represents the remaining number of undecayed atoms at time \(t\), \(N_0\) is the initial number of atoms, and \(e\) is the base of the natural logarithm. The equation can be differentiated to find the rate of decay events per second, providing insights into how active a sample is at any given moment. This rate helps in measuring the sample's radioactivity and planning for safe handling and disposal of radioactive materials. Using these calculations, students learn to quantify the microscopic processes occurring in radioactive samples.