Problem 58
Question
The half-life for the radioactive decay of \(\mathrm{C}-14\) is 5730 years and is inde- pendent of the initial concentration. How long does it take for 25\(\%\) of the C-14 atoms in a sample of \(C-14\) to decay? If a sample of C-14 initially contains 1.5 mmol of C-14, how many millimoles are left after 2255 years?
Step-by-Step Solution
Verified Answer
It takes 5730 years for 25% of C-14 to decay. After 2255 years, approximately 1.15 mmol of C-14 remains.
1Step 1: Understanding Half-Life
The half-life of a substance is the time it takes for half of the radioactive isotope to decay. In mathematical terms, after one half-life, the quantity of substance will be reduced to one-half of its initial value.
2Step 2: Calculate Time for 25% Decay
If it takes 5730 years for 50% to decay, it will take one more half-life for another half (25% of the original amount) to decay. Thus, it will be 75% decayed. Since we want to know when only 25% is decayed (75% remains), we calculate the time for 50% decay: one half-life or 5730 years. Therefore, in one half-life, 25% of the substance would have decayed.
3Step 3: Calculate Remaining C-14 After 2255 Years
To determine the remaining quantity of C-14 after a certain time, we use the exponential decay formula: \( N(t) = N_0(1/2)^{t/T} \), where \( N(t) \) is the quantity at time t, \( N_0 \) is the initial quantity, and T is the half-life. In this case, \( t = 2255 \) years and \( T = 5730 \) years. Substituting the values, we get \( N(2255) = 1.5(1/2)^{2255/5730} \).
4Step 4: Calculate the Exponent
Calculate the exponent value by dividing the given time by the half-life: \( 2255/5730 \) to determine the fraction of a half-life that has passed.
5Step 5: Compute Millimoles of C-14 Remaining
Plug the exponent value into the formula to find the remaining millimoles of C-14: \( N(2255) = 1.5(1/2)^{2255/5730} \). Use a calculator to compute the final value.
Key Concepts
Understanding Half-LifeExponential Decay FormulaIsotopesCarbon-14 Dating
Understanding Half-Life
Half-life is a crucial concept in the realm of nuclear physics, especially when discussing radioactive decay. Imagine a cluster of unstable radioactive atoms; the half-life is the time it takes for half of these atoms to transform into a more stable form through decay. In simpler terms, if you have 100 unstable atoms and the half-life is 1 year, you'd expect to have 50 left after 1 year, and then 25 after another year, and so forth.
This reduction continues in a predictable pattern, which is particularly important for applications such as carbon-14 dating. In our exercise, the half-life for carbon-14 (C-14) is given as 5730 years. That means if we start with a certain amount of C-14, we'd end up with half of that initial amount after 5730 years due to radioactive decay.
This reduction continues in a predictable pattern, which is particularly important for applications such as carbon-14 dating. In our exercise, the half-life for carbon-14 (C-14) is given as 5730 years. That means if we start with a certain amount of C-14, we'd end up with half of that initial amount after 5730 years due to radioactive decay.
Exponential Decay Formula
The exponential decay formula models the decrease of a substance over time. It takes the form of an exponential function, usually written as
\[ N(t) = N_0(1/2)^{t/T} \]
where \( N(t) \) represents the number of atoms remaining after time \( t \), \( N_0 \) is the original number of atoms, and \( T \) signifies the half-life of the substance. This formula is powerful because it describes how quantities diminish to half their size over and over again at regular intervals, hence the name 'exponential decay'. In the exercise, you see how this formula is used to calculate the remaining amount of carbon-14 after a specific period.
\[ N(t) = N_0(1/2)^{t/T} \]
where \( N(t) \) represents the number of atoms remaining after time \( t \), \( N_0 \) is the original number of atoms, and \( T \) signifies the half-life of the substance. This formula is powerful because it describes how quantities diminish to half their size over and over again at regular intervals, hence the name 'exponential decay'. In the exercise, you see how this formula is used to calculate the remaining amount of carbon-14 after a specific period.
Isotopes
Isotopes are variations of elements that have the same number of protons in each atom, but different numbers of neutrons. This subtle difference doesn't affect the chemical properties significantly, but it can change the nuclear stability of an atom. Radioactive isotopes, such as carbon-14, are isotopes that have an unstable nucleus and release energy to become more stable, a process known as radioactive decay. The number of neutrons in C-14, for instance, makes it radioactive, whereas the more common isotope carbon-12 is stable and doesn't undergo radioactive decay.
Carbon-14 Dating
Carbon-14 dating, also known as radiocarbon dating, is a method used by archaeologists and geologists to determine the age of organic materials. It's based on the principle that living organisms contain a certain amount of carbon-14. When the organism dies, it stops replenishing C-14, and the isotope begins to decay at a known rate characterized by its half-life. By measuring the ratio of C-14 to non-radioactive carbon isotopes in the sample and using the exponential decay formula, scientists can estimate the time since the organism's death. Thus, carbon-14 dating is a powerful tool for unraveling the history of ancient objects or remains.
Other exercises in this chapter
Problem 55
The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and has a rate constant of \(1.42 \times 10^{-4}
View solution Problem 57
The half-life for the radioactive decay of U- 238 is 4.5 billion years and is independent of initial concentration. How long will it take for 10\(\%\) of the U-
View solution Problem 60
A chemical reaction is endothermic and has an activation energy that is twice the value of the enthalpy change of the reaction. Draw a dia- gram depicting the e
View solution Problem 63
The rate constant \((k)\) for a reaction is measured as a function of tem- perature. A plot of ln \(k\) versus 1\(/ T(\) in \(\mathrm{K})\) is linear and has a
View solution