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Chapter 29: Problem 19

Many smoke detectors use small quantities of the isotope \({ }^{241} \mathrm{Am}\) in their operation. The half-life of \(2 \mathrm{t} \mathrm{Am}\) is 432 yr. How long will it take the activity of this material to decrease to \(1.00 \times 10^{-3}\) of the original activity?

Short Answer

Expert verified
It will take approximately 4303 years for the activity of this material to decrease to \(1.00 \times 10^{-3}\) of the original activity.

Step by step solution

01

Identify Given Values

First, identify the given values. We know the half-life, \(T = 432 yr\), and the final activity proportion, or \(\frac{N}{N_0} = 1.00 \times 10^{-3}\). Our goal is to find \(t\), the required time.
02

Express the Decay Formula

Using the decay formula \(N = N_0 \times(1/2)^{t/T}\) and given values, we obtain \((1.00 \times 10^{-3}) = (1/2)^{t/432}\) or equivalently, \(t/432 = log_{1/2} (1.00 \times 10^{-3})\). This turns our exponential equation into a logarithmic equation that can be more easily solved for \(t\).
03

Solve for Time (\(t\))

Next, solve for time. The log base \(1/2\) calculation can be rewritten as \(-log_{2} (1.00 \times 10^{-3})\), which calculates to approximately 9.97. So \(t/432 = 9.97\). Solving for \(t\) results in \(t \approx 4303 yr\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Life
Understanding the half-life of a radioactive isotope is crucial in various scientific fields, including geology, archeology, and medicine. The half-life is the time required for half of a sample of a radioactive isotope to decay into a more stable form.

In simple terms, think of it as a clock that tells us how 'old' or 'new' a radioactive material is, based on how much of it has decayed. If we have a 100g sample of a substance with a half-life of 10 years, in 10 years we'll have 50g of the original substance, and after another 10 years, only 25g will remain. This process is exponential, meaning it doesn't happen at a steady rate, but rather, it slows down as time passes.
Isotopes
Isotopes are versions of a chemical element that have the same number of protons but different numbers of neutrons. This means while they share chemical properties, their atomic mass and nuclear stability vary.

Some isotopes, like the ones used in smoke detectors, are radioactive. The isotope Americium-241 mentioned in the exercise is one such example. It emits radiation until it transforms into a more stable isotope. These properties make certain isotopes useful in a variety of applications: from dating archaeological finds (using isotopes like Carbon-14) to treating cancer (using isotopes like Cobalt-60).
Exponential Decay
Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value. In the context of radioactive decay, this means the amount of a radioactive isotope diminishes exponentially over time.

The decay formula, often expressed as \( N = N_0 \times(1/2)^{t/T} \), captures this concept. Here, \( N \) is the remaining quantity after time \( t \), \( N_0 \) is the original quantity, and \( T \) is the half-life. It's the 'exponential' part of exponential decay that ensures the substance doesn't vanish all at once but rather fades away gradually, mirroring the half-life property.
Logarithmic Equations
Logarithmic equations are the inverse of exponential equations and can be particularly useful when determining the time frame for processes like radioactive decay. We often come across them when we're trying to find out 'when' rather than 'how much'.

By rearranging the decay formula to solve for time, we're essentially taking the logarithm of both sides. This truncates the exponential process into a more familiar linear form. In the exercise's solution, the logarithmic equation allowed us to calculate time without directly dealing with the rapidly decreasing quantity of the substance.

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Most popular questions from this chapter

The half-life of an isotope of phosphorus is 14 days. If a sample contains \(3,0 \times 10^{16}\) such nuclei, determine its activity. Express your answer in curies.

Using \(2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3}\) as the density of nuclear matter, find the radius of a sphere of such matter that would have a mass equal to that of Earth. Earth has a mass equal to \(5.98 \times 10^{24} \mathrm{~kg}\) and average radius of \(6.37 \times 10^{6} \mathrm{~m}\).

Determine which of the following suggested decays can oceur spontaneously: (a) \({ }_{20}^{+0} \mathrm{Ca} \rightarrow \mathrm{e}^{+}+{ }_{19}^{40} \mathrm{~K}\) (b) \(_{\text {in }}^{144} \mathrm{Nd} \rightarrow{ }_{2}^{4} \mathrm{He}+{ }_{5 \times}^{140} \mathrm{Ce}\)

A particular radioactive source produces \(100 \mathrm{mrad}\) of 2-MeV gamma rays per hour at a distance of \(1.0 \mathrm{~m}\). (a) How long could a person stand at this distance before accumulating an intolerable dose of 1 rem? (b) Assuming the gamma radiation is emitted uniformly in all directions, at what distance would a person receive a dose of \(10 \mathrm{mrad} / \mathrm{h}\) from this source?

A patient swallows a radiopharmaceutical tagged with phosphorus-32 \(\left(\begin{array}{c}32 \\ 15\end{array}\right)\), a \(\beta^{-}\) emitter with a half-life of \(14.3\) days. The average kinetic energy of the emitted electrons is \(700 \mathrm{keV}\). If the initial activity of the sample is 1.31 MBq, determine (a) the number of electrons emitted in a 10 -day period, (b) the total energy deposited in the body during the 10 days, and (c) the absorbed dose if the electrons are completely absorbed in \(100 \mathrm{~g}\) of tissue.
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