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 $64.0.$ 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 $25\mathrm{\%}$ of the initial measured value. How long does she have to do the experiment?

Short answer

a. The decay constant, $k=6.265\times {10}^{-7}{\mathrm{s}}^{-1}$. b. The number of decay events in the first second of a 28.0 mg sample of copper-64 is $1.66\times {10}^{15}decayevents$. c. The chemist has 30.5 days to do the experiment before the radioactivity falls below 25% of the initial value.

Step-by-step solution

Calculate the decay constant

We need to find the decay constant, k in s^-1. We can use the formula for half-life: $t_{1/2}=\frac{ln(2)}{k}$, where $t_{1/2}$ is the half-life of the element. Rearrange the formula to get: $k=\frac{ln(2)}{t_{1/2}}$ First, we will need to convert half-life '12.8 days' to seconds: $12.8days\times 24hours/day\times 60minutes/hour\times 60s/minute=1105920s$ Now, plug in the value of $t_{1/2}$ into the formula: $k=\frac{ln(2)}{1105920s}=6.265\times {10}^{-7}{\mathrm{s}}^{-1}$

Calculate the number of decay events in the first second

We have a sample of 28.0 mg of copper-64, and its atomic mass is 64.0. First, we need to find the number of atoms, N, from the given mass: Formula: $N=\frac{mass}{atomicmass}\times N_A$, where $N_A$ is Avogadro's number, $6.022\times {10}^{23}atoms/mol$. Plug in the values for the given mass and atomic mass of copper-64: $N=\frac{28.0\mathrm{mg}}{64.0g/mol}\times 6.022\times {10}^{23}atoms/mol$ $N=2.65\times {10}^{21}atoms$ Now we can use the decay constant, k, and the number of atoms, N, to find the number of decay events in the first second. The decay rate, R, at any given time is given by the formula: $R(t)=-\frac{dN(t)}{dt}=kN(t)$ We are asked for the decay rate at the first second, so t = 1 second. At t = 0 seconds, N(t) = N (the initial number of atoms). Then, R(1 s) = kN: $R(1s)=6.265\times {10}^{-7}{\mathrm{s}}^{-1}\times 2.65\times {10}^{21}atoms=1.66\times {10}^{15}decayevents$

Calculate the time available for the experiment

To find the time available for the experiment, we need to solve the radioactive decay formula $N(t)=N(0)e^{-kt}$ for t, given that N(t) cannot fall below 25% of N(0). Let's set N(t) = 0.25N(0) and solve for t: $0.25N(0)=N(0)e^{-kt}$ Divide both sides by $N(0)$: $0.25=e^{-kt}$ Now, take the natural logarithm of both sides: $ln(0.25)=-kt$ Finally, solve for t: $t=-\frac{ln(0.25)}{k}=\frac{ln(4)}{6.265\times {10}^{-7}{\mathrm{s}}^{-1}}=2.638\times {10}^{6}s$ Now let's convert the seconds to days: $2.638\times {10}^{6}s\times \frac{1min}{60s}\times \frac{1hour}{60min}\times \frac{1day}{24hours}=30.5days$ The chemist has 30.5 days to do the experiment before the radioactivity falls below 25% of the initial value.

Key concepts

Half-life

Half-life is a term used to describe the time it takes for half of the radioactive atoms in a sample to decay. This concept is crucial in understanding radioactive decay because it allows scientists and researchers to predict how long a radioactive substance will remain active. A half-life is characteristic of each radioactive isotope and does not depend on the initial amount of the substance or its concentration. In simpler terms, if you have a sample of a radioactive element with a half-life of 5 years, after 5 years only half of the sample will remain; the other half will have transmuted into different atoms or energy. Half-life allows for the calculation of the decay constant, which is necessary for solving various nuclear chemistry problems.

When improving educational material on half-life, it's important to ensure that students can connect the concept to practical examples, such as medical imaging with radioactive tracers or carbon dating in archaeology, which helps in reinforcing the concept's real-world application.

Decay Constant

The decay constant, represented by the symbol $k$, is a probability that quantifies the rate at which an atom of a radioactive substance decays per unit time. Mathematically, the decay constant is the inverse of the half-life multiplied by the natural logarithm of 2 ($ln(2)$). The formula $k=\frac{ln(2)}{t_{1/2}}$ directly relates the decay constant to half-life, showing that a substance with a shorter half-life has a higher decay constant, indicating it is more radioactive and will decay faster.

To make the concept more understandable in educational content, it's useful to incorporate visual aids, like decay curves, that illustrate how the probability of decay changes over time. Additionally, providing step-by-step calculations, as seen in the original exercise, can help students grasp the process of determining the decay constant from a known half-life.

Nuclear Chemistry

Nuclear chemistry is the subfield of chemistry dealing with radioactivity, nuclear processes, and nuclear properties. It involves studying various types of radioactive decay, nuclear synthesis, and the chemical applications of radioisotopes. A common application of nuclear chemistry is in medicine, where isotopes are used for diagnosis and treatment in nuclear medicine. Another pivotal application is in the field of energy production through nuclear reactors. Understanding the basics of nuclear transformations, such as alpha decay, beta decay, and gamma radiation, is essential for those studying nuclear chemistry.

In educational content, explaining the interplay between mass, energy, and the stability of the nucleus aids in the comprehensive understanding of nuclear chemistry. Highlighting safety precautions and ethical considerations is also important due to the potential risks associated with radioactive materials. Introducing historical context, such as the development of the atomic theory and notable discoveries, can engage students and provide a broader perspective on the subject.