Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The principle behind the home smoke detector is described on page 516. Americium- 241 is present in such detectors. It has a decay constant of \(1.51 \times 10^{-3} \mathrm{y}^{-1}\). You are urged to check the battery in the detector at least once a year. You are, however, never encouraged to check how much Am-241 remains undecayed. Explain why.

Short Answer

Expert verified
One is encouraged to check the battery in a smoke detector because the battery's lifespan is much shorter than the half-life of Am-241, which is approximately 459 years. This means that the Am-241 remains effective for a long time and does not lose its functionality significantly over a human's typical lifespan. Therefore, it is important to check the battery annually to ensure the proper functionality of the smoke detector, while there is no need to check the remaining amount of Am-241 due to its long half-life.

Step by step solution

01

Find the half-life of Am-241

To find the half-life of Am-241, we can use the decay constant given in the problem: \(\lambda = 1.51 \times 10^{-3} \mathrm{y}^{-1}\). The relation between half-life, T, and decay constant is given by the formula: \(T = \frac{\ln 2}{\lambda}\). Substituting the given decay constant, we can calculate the half-life of Am-241. \(T = \frac{\ln 2}{1.51 \times 10^{-3}} = \frac{0.693}{1.51 \times 10^{-3}} \approx 459 \,\text{years}\)
02

Compare the half-life to the advised checking frequency

The half-life of Am-241 is approximately 459 years, which is significantly larger than the recommended checking frequency for the battery, which is once a year. This means that the Am-241 material in the smoke detector will not decay substantially over a human's typical lifespan.
03

Discuss the implications of a large half-life and decay constant

The activity of Am-241 is governed by the decay constant, and since it has a small decay constant and a large half-life, that means it will remain effective for a long time. Since it doesn't decay substantially, there would be no need to check how much Am-241 remains undecayed in the smoke detector regularly.
04

Concluding remarks

In conclusion, the reason one is not encouraged to check the remaining amount of Am-241 is that it has a long half-life, meaning it doesn't lose its effectiveness significantly over time. Instead, the battery of the home smoke detector is more likely to run out before the Am-241 becomes ineffective, so checking the battery annually is more important to ensure the proper functionality of the smoke detector.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding Half-Life
Half-life is an essential concept in radioactive decay. It represents the time required for half of the radioactive nuclei in a sample to decay. This doesn't mean that after one half-life all of the nuclei are gone; instead, the process is exponential. Each half-life results in half of the remaining radioactive atoms decaying, but the other half stays intact.

For example, if you start with 100 nuclei, after one half-life, 50 would remain. After two half-lives, 25 would remain, and so on. It’s a continuous and ongoing process. This characteristic makes half-life a reliable way to predict how long a radioactive material will remain active and how it will decrease over time.

Half-life is calculated using the decay constant, with the formula:
  • \[T = \frac{\ln 2}{\lambda}\]
where \(T\) is the half-life and \(\lambda\) is the decay constant.

The half-life of a material such as Americium-241 is crucial because it helps us understand the timeframe over which the material will remain active, like in smoke detectors.
Role of Decay Constant
The decay constant is a unique value that indicates the probability of a radioactive atom decaying per unit time. It essentially measures how quickly or slowly a radioactive isotope undergoes decay.

A higher decay constant means the material will decay more rapidly, while a lower decay constant indicates a slower rate of decay. The decay constant (\(\lambda\)) and half-life (\(T\)) are inversely related, which means if the decay constant is small, the half-life is long and vice-versa.

For Americium-241, the decay constant is \(1.51 \times 10^{-3} \, \text{y}^{-1}\), implying it decays slowly. That slow rate is why the half-life of Am-241 is long, specifically around 459 years. This slow decay rate ensures that the Americium-241 used in smoke detectors remains active for many years, providing safety and reliability without needing frequent checks.
Americium-241 in Smoke Detectors
Americium-241 is a radioactive isotope used in many smoke detectors due to its long half-life and stable decay characteristics.

Why is it used? Here are some reasons:
  • It emits alpha particles that can ionize air, allowing smoke detectors to detect smoke particles effectively.
  • Its long half-life of approximately 459 years ensures that it remains functional over the typical lifespan of a smoke detector.
  • Am-241 decays slowly, making it safe for household use, as the risk of radiation exposure is minimal.
Given these characteristics, once Americium-241 is installed in a smoke detector, users do not need to worry about it becoming ineffective anytime soon. The device will likely wear out due to other components, such as the battery running out, long before the Americium decays significantly. This is why regular battery checks are crucial, ensuring that the smoke detector remains operational.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An explosion used five tons \((1\) ton \(=2000 \mathrm{lb}\) ) of ammonium nitrate \((\Delta E=-37.0 \mathrm{~kJ} / \mathrm{mol})\). (a) How much energy was released by the explosion? (b) How many grams of TNT \((\Delta E=-2.76 \mathrm{~kJ} / \mathrm{g})\) are needed to release the energy calculated in (a)? (c) How many grams of U-235 are needed to obtain the same amount of energy calculated in (a)? (See the equation in Problem 45.)

Cobalt-60 is used extensively in medicine as a source of \(\gamma\) -rays. Its half-life is \(5.27\) years. (a) How long will it take a \(\mathrm{Co}-60\) source to decrease to \(18 \%\) of its original activity? (b) What percent of its activity remains after 29 months?

Radium-226 decays by alpha emission to radon-222. Suppose that \(25.0 \%\) of the energy given off by one gram of radium is converted to electrical energy. What is the minimum mass of lithium that would be needed for the voltaic cell \(\mathrm{Li}\left|\mathrm{Li}^{+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\), at standard conditions, to produce the same amount of electrical work \(\left(\Delta G^{\circ}\right) ?\)

Consider the fission reaction in which U-235 is bombarded by neutrons. The products of the bombardment are rubidium-89, cerium-144, beta particles, and more neutrons. (a) Write a balanced nuclear equation for the bombardment. (b) Calculate \(\Delta E\) when one gram of U-235 undergoes fission. (c) The detonation of TNT, an explosive, evolves \(2.76 \mathrm{~kJ} / \mathrm{g}\). How many kilograms of TNT are required to produce the same amount of energy as one milligram of \(\mathrm{U}-235\) ?

Which has the larger binding energy, fluorine- 19 or oxygen-17?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free