The Chemistry of Smoke Detectors
- Page ID
- 418931
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- I. F. 2. Atoms change identity in nuclear reactions; it is possible to write nuclear equations that follow these changes.
- a. Balancing nuclear reactions is based on the simultaneous conservation of both atomic and mass numbers.
- VII. B. 1. The “order” of a reaction is derived from the exponent on the concentration term of a given reactant in the rate law.
- c. The kinetics of nuclear processes is first order and can be quantitatively treated, often by indicating the half-life of a nuclear reaction.
Introduction
Radioactive decay is the process in which radioactive materials break down. Oftentimes, a radioactive element or an isotope’s half-life, or the amount of time it takes for half of the original sample to decompose, is used to determine how quickly a sample decomposes in comparison to other radioactive material. Radioactive materials and their decay have proven to be useful in many fields of science, such as creating nuclear weapons and even creating a source of sustainable energy. However, another unique and unexpected way in which radioactive decay is used in real life is in smoke detectors.
Background Information
Types of Decay
There are three main types of radioactive decay: alpha decay, beta decay, and positron emission. Each type of decay is classified based on what particle it emits and the change in the original isotope’s atomic and mass numbers.
Alpha decay is popular in large unstable isotopes as it changes to another element that has two less protons by emitting an alpha particle (\({}_{2}^{4} \alpha\)) composed of two protons and two neutrons. These particles have large masses and positive charge. An example is shown below:
\[{}_{88}^{226} Ra \rightarrow{}_{86}^{222} Rn+{}_{2}^{4} \alpha\]
Beta decay is characterized by the emission of a beta particle (\({}_{-1}^{ 0} e\)) and the change of the original element into another element that has one more proton; in other words, the new element has an atomic number that is one more than the original atomic number.
\[{}_{6}^{14} C \rightarrow{}_{7}^{14} N+{}_{-1}^{0} e\]Positron emission is characterized by the emission of a positron (\({}_{+1}^{ 0} e\)) and the change of the original element into another element that has one less proton; in other words, the new element has an atomic number that is one less than the original atomic number.
\[{}_{8}^{15} O \rightarrow{}_{7}^{15} N+{}_{+1}^{0} e\]
Half-Life Equations
Radioactive decay is an example of a first order reaction. Each type of radioactive decay reaction is a first order reaction, which means that the rate of the reaction only depends on the concentration of one reactant. In this case, that one reactant is the radioactive nuclei, and in other words, the rate of reaction depends only on the concentration of the nuclei. Because the rate at which certain isotopes decay is determined by the half-life of that isotope, it is important to derive an equation that can tell us what the half-life of an isotope is, given that we know the rate constant of the decay reaction. The derivation for the half-life of an isotope in a first order reaction is as follows:
\[[A]_t=[A]_0 e^{-k t}\nonumber\]
We can then take the natural logarithm of both sides.
\[ln \frac{[A]_0}{[A]_t}=k t\nonumber\]
We can rearrange the equation using logarithmic laws to turn it into a more familiar equation.
\[ln{[A]_t} - ln{[A]_0}=-k t\nonumber\]
Based on logarithmic laws, we can rewrite the left side of the equation. Recall that the concentration of a substance at its half life is simply half of the original concentration; based on this, we can make a substitution. The initial concentration values cancel out.
\[ln \frac{0.5 [A]_0}{[A]_0}=-k t_{1/2}\nonumber\]
Here, we are left with ln(0.5). We can simplify that value to -0.693.
\[ln{(0.5)}=-k t_{1/2}\nonumber\]
Then, after dividing by the rate constant and by -1 on both sides, we get our equation for the half-life of an isotope in a first
order reaction.
\[\frac{0.693}{k}=t_{1/2}\]
Application of Smoke Detectors
Ever since 1976, in accordance with the Life Safety Code’s requirements, many indoor spaces in the U.S. have had one thing in common: the presence of small, disc-like objects known as smoke detectors. Smoke detectors are essential for households, as they detect smoke and even carbon monoxide from sources that may not be visible to the eye. Interestingly, smoke detectors depend on radioactivity to function. Most detectors use americium-241 as the source. Via an alpha-decay reaction, americium-241 produces neptunium-237 and an alpha particle, as shown below:
\[{}_{95}^{241} Am \rightarrow{}_{93}^{237} Np+{}_{2}^{4} \alpha\]
These alpha particles remove electrons from neutral air molecules, creating positive ions. The now-removed electrons attach to other neutral air molecules, creating negative ions. In the detector, two charged plates attract the positive and negative ions, creating a weak current, which can be seen below.
Figure 1. Diagram of Smoke Detector Before Presence of Smoke (from U.S. Nuclear Regulatory Commission)
When smoke particles enter into the chamber, they will interrupt this flow; it is when the current is broken that the smoke detector goes off.
Figure 2. Diagram of Smoke Detector After Presence of Smoke (from U.S. Nuclear Regulatory Commission)
As shown in the image above, the typical attraction between the ions and the charged plates is interrupted by the smoke particles. When the current drops below a certain threshold, the familiar, screeching sound of a smoke detector goes off.
Common Questions
How often do I need to replace my smoke detectors?
According to the CDC, roughly 0.0002 g of americium-241 is present in each smoke detector, and according to the National Fire Protection Association, a smoke detector should be replaced every 10 years. Can this be accredited to the small sample of americium-241 decaying almost completely after 10 years?
Because radioactive samples won’t decay completely, find out how long it takes for there to be 0.00001% of the sample left given that americium-241 has a half-life of 432 years.
Solution
We start with the following equation:
\[{\dfrac{1}{2}}^x=0.0000001\nonumber\]
This equation is the typical half-life equation, where x represents the amount of half lives a substance goes through, and the number on the other side of the equation represents the proportion of the original sample left. We can verify that this equation is true because if x is equal to 1 half life, then the number on the other side of the equation will be 0.5. From here, we can take the log base 0.5 of both sides; the left side will cancel out and will leave just x.
\[x=log_{0.5}{(0.0000001)}\nonumber\]
Putting this equation into a calculator, x equates to 23.25, meaning for there to be 0.00001% of the sample left, the substance must go through 23.25 half lives. We know that the half life of americium-241 is 432 years, so we multiply the amount of half lives we have (23.25) by the half life of americium-241 (432 years).
\[23.25 * 423 = 10,044 years\nonumber\]
Because it takes more than 10,044 years for 0.0002 g of americium-241 to almost completely decay, we can assume that it is not because smoke detectors run out of radioactive material that we have to replace them every 10 years. Instead, over time, dust and other particles tend to accumulate in the smoke detectors, which makes them less effective as time goes on.
Should I be worried about radioactive materials in my house?
By including radioactive materials in the design of smoke detectors, it poses a risk to human health, as it can damage the DNA of our cells if exposed too long. Let's perform calculations to determine how much radioactive material is present in the atmosphere around a smoke detector after some amount of time.
Given that a brand new smoke detector has 0.33μg of americium-241, how many grams of the produced isotope will be left after two years have elapsed? Recall that the half-life of americium-241 is 432 years.
Solution
To find the number of half-lives of americium-241 undergone in two years, we do:
\[\dfrac{2}{432} = 0.00463 years\nonumber\]
To calculate the portion of substance remaining, we do:
\[{\dfrac{1}{2}}^{0.00463}={0.9968}\nonumber\]
Therefore, 99.68% of americium-241 remains after two years. Consequently, 0.32% of americium-241 is used up after two years. Let's quickly convert from micrograms to grams with the given micrograms.
\[\dfrac{0.33 g}{10^{-6} μg} = {3.3 × 10^{-7} g}\nonumber\]
Now, to find the amount of grams after two years, we do:
\[{0.0032}{⋅}{3.3 × 10^{-7} g} = {1.056 × 10^{-9} g {}^{241}Am}\nonumber\]
In reference to Equation 5 (see above), 1 mole of americium-241 is equivalent to 1 mole of neptunium-237. Therefore, 1.056 × \(10^{-9}\) g \({}^{237}Np\) is produced after 2 years.
Reflecting on this answer, there will be a miniscule amount of neptunium-237 produced from the smoke detector. This amount will not impact a human's health, so there is nothing to worry about in terms of radioactivity.
References
(1) Backgrounder on smoke detectors | nrc.gov. https://www.nrc.gov/reading-rm/doc-c...detectors.html.
(2)Eason, E. Americium Smoke Detectors. http://large.stanford.edu/courses/2011/ph241/eason1/.
(3) Americium. https://wwwn.cdc.gov/TSP/PHS/PHS.asp...ke%20detectors.
(4) Odonnell, F. R.; Etnier, E. L.; Holton, G. A.; Travis, C. C. Assessment of radiation doses from residential smoke detectors that contain americium-241. https://ui.adsabs.harvard.edu/abs/19.......O/abstract.
(5) Xing, D.; Liang, C.; Zhang, X. An 241Am Plasma Desorption Ionization (AmDI) Source Scavenged from Smoke Detectors for Ambient Mass Spectrometry Sampling. https://pubs.acs.org/doi/10.1021/acs.analchem.2c01704.