Problem 67 - Chemistry: The Central Science

Question

Americium-241 is an alpha emitter used in smoke detectors. The alpha radiation ionizes molecules in an air-filled gap between two electrodes in the smoke detector, leading to current. When smoke is present, the ionized molecules bind to smoke particles and the current decreases; when the current is reduced sufficiently, an alarm sounds. (a) Write the nuclear equation corresponding to the alpha decay of americium-241. (b) Why is an alpha emitter a better choice than a gamma emitter for a smoke detector? (c) In a commercial smoke detector, only \(0.2\) micrograms of americium are present. Calculate the energy that is equivalent to the mass loss of this amount of americium due to alpha radiation. The atomic mass of americium- 241 is \(241.056829\) amu. (d) The half-life of americium- 241 is 432 years; the half life of americium-240 is \(2.12\) days. Why is the 241 isotope a better choice for a smoke detector?

Step-by-Step Solution

Answer

(a) The nuclear equation for the alpha decay of americium-241 is: ^{241}Am -> ^{4}He + ^{237}Np. (b) Alpha emitters are better for smoke detectors because they have high ionizing power but low penetration power, making them more effective in detecting changes in current due to smoke. (c) The energy equivalent to the mass loss of 0.2 micrograms of americium due to alpha radiation is 45.0405 J/kg. (d) Americium-241 is a better choice for smoke detectors because of its long half-life of 432 years, ensuring continuous and consistent ionization for many years, while americium-240 has a short half-life of 2.12 days, requiring frequent replacement.

Step 1: Nuclear Reaction for Alpha Decay of Americium-241

Alpha decay occurs when a nucleus emits an alpha particle, which consists of two protons and two neutrons. The Americium-241 (^{241}Am) nucleus undergoes alpha decay, and we need to find the decay product: ^{241}Am -> ^{4}He + X, where X is the nucleus formed after alpha decay. As two protons and two neutrons are emitted as the alpha particle, the new nucleus would have 2 fewer protons and 2 fewer neutrons. Americium has an atomic number of 95, so the new nucleus (X) will have an atomic number of 93. The mass number of the new nucleus will be 241 - 4 = 237. This corresponds to element neptunium (^{237}Np): ^{241}Am -> ^{4}He + ^{237}Np.

Step 2: Explain why alpha emitter is better than gamma emitter in a smoke detector

Alpha particles (alpha radiation) have high ionization power but low penetration power since they are relatively large, heavy particles. They can easily ionize molecules in the air but can't penetrate materials as easily as gamma radiation. Due to their ionizing power, alpha particles are a better choice for smoke detectors because they are more likely to cause changes in the current that can be detected when smoke is present. On the other hand, gamma radiation has high penetration power and low ionization power due to its nature as a high-energy photon. Gamma radiation would pass through most materials, making it ineffective for smoke detectors because it would not create significant changes in the current to detect the presence of smoke.

Step 3: Calculate energy equivalent to mass loss of 0.2 micrograms of Americium

First, convert the mass of americium from micrograms to kilograms: \(0.2\,\text{μg} = 2 \cdot 10^{-10}\,\text{kg}\). Now we need to find the mass loss during alpha decay, which corresponds to the mass difference between the initial americium-241 nucleus and the final products (alpha particle and neptunium-237 nucleus). Since the nucleus is relatively stable, we can use atomic masses as an approximation of nuclear masses. Subtract the atomic mass of ^{4}He (4.002602 amu) and ^{237}Np (237.0482 amu) from the atomic mass of ^{241}Am (241.056829 amu): mass loss = \(241.056829 - (4.002602 + 237.0482) = 0.006027\,\text{amu}\) To find the mass loss in kilograms, multiply by the conversion factor (1 amu = 1.660539 \cdot 10^{-27} kg): mass loss = \(0.006027 \cdot 1.660539 \cdot 10^{-27}\,\text{kg/amu} = 1.0009 \cdot 10^{-29}\,\text{kg}\) Now use Einstein's mass-energy equivalence formula, E=mc^2, to find the energy equivalent of this mass loss: E = (1.0009 \cdot 10^{-29}\,\text{kg}) \cdot (3 \cdot 10^8\,\text{m/s})^2 = 9.0081 \cdot 10^{-12}\,\text{J}$ Since smoke detectors use only 0.2 micrograms of americium, we need to find the energy equivalent per decay by dividing the energy obtained above by the initial mass: Energy per decay = \(\frac{9.0081\cdot 10^{-12}}{2\cdot 10^{-10}} = 45.0405\,\text{J/kg}\)

Step 4: Compare the half-life of Americium-241 and Americium-240 and choose the better isotope for smoke detectors

The half-life of americium-241 is 432 years, while the half-life of americium-240 is 2.12 days. A long half-life is important for smoke detector applications because it ensures the device will remain functional for a long period without needing to replace the radioactive source. In this case, americium-241, with its much longer half-life, is a better choice for smoke detectors as it will provide continuous and consistent ionization for many years, ensuring the smoke detector remains functional and effective. Americium-240, on the other hand, would require frequent replacement of the radioactive source, which is not practical for smoke detectors.

Key Concepts

Alpha Decay

Alpha decay plays a crucial role in a vast array of applications, from the dating of archaeological finds to its use in smoke detectors. It occurs when an unstable nucleus emits an alpha particle, consisting of two protons and two neutrons, effectively becoming a new element. For instance, when americium-241 undergoes alpha decay, it transforms into neptunium-237:
\(^{241}_{95}Am \rightarrow ^{4}_{2}He + ^{237}_{93}Np\)
This nuclear reaction is accompanied by a slight loss in mass and a release of energy, phenomena that are cornerstones of nuclear chemistry.

Smoke Detectors

The practical usage of alpha decay is epitomized by its application in smoke detectors, a critical safety device found in homes and businesses. In these devices, alpha particles are utilized to ionize air molecules between two electrodes, creating a current. When smoke enters, it disrupts this current by attaching to the ionized particles, resulting in the decrease of the current and consequently, triggering the alarm. The choice of an alpha emitter in smoke detectors hinges on its high ionization power yet low penetration, which ensures it does not escape the device but is sufficiently sensitive to the presence of smoke.

Why Alpha and Not Gamma?

Gamma emitters are less suited for smoke detectors due to their high penetration power, meaning they would not effectively interact with the air to produce a detectable current.

Half-Life

The half-life of a radioactive isotope is a term that defines the time required for half of the isotope's atoms to decay. Understanding half-life is vital for a variety of applications, including medical treatments and archaeological dating, as well as for selecting the appropriate isotope for smoke detectors. In this context, americium-241, with its half-life of 432 years, provides a long period of activity compared to americium-240's 2.12 days, making it far more stable and suitable for the long-term use required by smoke detectors. The longer half-life translates to less frequent replacements, enhancing the smoke detector's reliability and safety.

Mass-Energy Equivalence

Mass-energy equivalence, encapsulated by Einstein’s famous equation \(E = mc^2\), describes the principle that mass can be converted into energy and vice versa. In alpha decay, a small amount of mass is transformed into energy. For instance, the minuscule mass loss of 0.2 micrograms of americium due to alpha radiation amounts to substantial energy. This conversion is what fuels the decay process, and understanding this phenomenon is key for applications such as assessing the energy output in nuclear reactors or, more specifically, determining the operational lifespan of a smoke detector's radioactive source.

Nuclear Equation

Nuclear equations are used to represent the transformations that occur during nuclear reactions, such as alpha decay. They closely resemble chemical equations but highlight changes in atomic and mass numbers. A nuclear equation must be balanced, which means that the numbers of protons and neutrons are conserved through the reaction. In the provided exercise, the equation accurately indicates the conversion of americium-241 into neptunium-237 and an alpha particle while adhering to the law of conservation of mass and charge. Writing and understanding these equations is fundamental to the study of nuclear chemistry and helps illustrate the changes occurring within nuclear reactions.