Problem 1
Question
Smoke detectors contain a small amount of americium-241. Its decay product is neptunium-237. Identify the emission from americium-241.
Step-by-Step Solution
Verified Answer
Answer: Americium-241 decays to neptunium-237 through the emission of an alpha particle.
1Step 1: Identifying the parent nucleus (americium-241) and daughter nucleus (neptunium-237)
The parent nucleus in this decay process is americium-241. Its symbol is represented by Am. The daughter nucleus is neptunium-237, represented by Np.
2Step 2: Finding the atomic numbers and mass numbers of the elements
The atomic number of americium is 95, and the atomic number of neptunium is 93. The mass number of Am-241 is 241, and the mass number of Np-237 is 237.
3Step 3: Using the conservation of nucleon numbers to find the missing particle
The decay process conserves both the total atomic number (proton number) and the total mass number (proton number + neutron number). We can write two conservation equations:
Atomic number conservation: \(Z_{Am} - Z_{Np} = Z_{emission}\)
Mass number conservation: \(A_{Am} - A_{Np} = A_{emission}\)
Plugging in the values, we can find the atomic number (Z) and mass number (A) of the emitted particle:
\(95 - 93 = Z_{emission}\) → \(Z_{emission} = 2\)
\(241 - 237 = A_{emission}\) → \(A_{emission} = 4\)
4Step 4: Identifying the emitted particle
Based on the calculations above, the emitted particle has an atomic number of 2 and a mass number of 4. This corresponds to an alpha particle (\(\alpha\)), which consists of 2 protons and 2 neutrons. Therefore, the emission from americium-241 is an alpha particle.
Key Concepts
Americium-241 DecayNeptunium-237Conservation of Nucleon NumbersRadioactive DecayNuclear Chemistry
Americium-241 Decay
Americium-241 is a remarkable element widely used in smoke detectors for its ability to detect smoke through a process known as alpha decay. When an Americium-241 atom decays, it releases an alpha particle, which comprises two protons and two neutrons. This release results in the transformation of the original atom into another element, Neptunium-237.
An alpha particle is a helium nucleus, and since it contains charged particles, its emission can ionize air molecules. In the context of a smoke detector, alpha particles collide with air molecules and free electrons, creating ions. If smoke enters the detector, these particles attach to the ions, disrupting the current and triggering the alarm. This practical application of nuclear chemistry provides a vital safety mechanism in households and buildings around the world.
An alpha particle is a helium nucleus, and since it contains charged particles, its emission can ionize air molecules. In the context of a smoke detector, alpha particles collide with air molecules and free electrons, creating ions. If smoke enters the detector, these particles attach to the ions, disrupting the current and triggering the alarm. This practical application of nuclear chemistry provides a vital safety mechanism in households and buildings around the world.
Neptunium-237
Following the decay of Americium-241, Neptunium-237 emerges. This new element has two fewer protons than Americium-241, which is precisely the atomic make-up of an alpha particle. Neptunium-237 is a less well-known element due to its scarcity and origins primarily from the decay of other radioactive materials. As a synthetic element with a long half-life, it further undergoes radioactive decay, contributing to the chain of transformations in nuclear chemistry.
Despite its obscurity, Neptunium-237 has potential uses in the field of nuclear science and energy. For instance, it has been considered as a possible material for nuclear reactors and even the production of nuclear weapons, reflecting the complexity and range of elements produced through radioactive decay.
Despite its obscurity, Neptunium-237 has potential uses in the field of nuclear science and energy. For instance, it has been considered as a possible material for nuclear reactors and even the production of nuclear weapons, reflecting the complexity and range of elements produced through radioactive decay.
Conservation of Nucleon Numbers
When examining nuclear reactions, such as alpha decay, the principle of the conservation of nucleon numbers is essential. This principle states that the sum of protons and neutrons, known as nucleons, remains constant before and after a nuclear reaction. Through this conservation, we can trace the changes that occur within an atom's nucleus.
The total number of protons, which defines the atomic number, and the total number of nucleons, reflecting the mass number, should equal those of the products after the decay. In our case with Americium-241 decaying into Neptunium-237 plus an alpha particle, the atomic numbers and mass numbers on both sides of the equation match, satisfying the conservation laws. These principles are fundamental to understanding nuclear reactions and ensure that every particle is accounted for in the transformation process.
The total number of protons, which defines the atomic number, and the total number of nucleons, reflecting the mass number, should equal those of the products after the decay. In our case with Americium-241 decaying into Neptunium-237 plus an alpha particle, the atomic numbers and mass numbers on both sides of the equation match, satisfying the conservation laws. These principles are fundamental to understanding nuclear reactions and ensure that every particle is accounted for in the transformation process.
Radioactive Decay
Radioactive decay is the process by which an unstable atomic nucleus loses energy through the emission of radiation. There are several types of decay, with alpha decay being just one. In alpha decay, an atom releases an alpha particle, significantly altering its chemical identity as it transforms into a different element.
Radioactive decay is a spontaneous process and is not influenced by external factors such as temperature or pressure. It's characterized by the half-life, which is the time required for half of a sample of a radioactive substance to decay. The study of these processes falls under the category of nuclear chemistry, as it involves changes in atomic nuclei, differing from traditional reactions that involve the exchanging of electrons.
Radioactive decay is a spontaneous process and is not influenced by external factors such as temperature or pressure. It's characterized by the half-life, which is the time required for half of a sample of a radioactive substance to decay. The study of these processes falls under the category of nuclear chemistry, as it involves changes in atomic nuclei, differing from traditional reactions that involve the exchanging of electrons.
Nuclear Chemistry
Nuclear chemistry is a specialized field that focuses on the reactions, transformations, and behaviours of atomic nuclei. It encompasses not only radioactive decay but also fission, fusion, and the synthesis of new elements. Nuclear processes power stars, including our sun, and are harnessed for energy on Earth through nuclear reactors.
Understanding nuclear chemistry is vital for multiple real-world applications ranging from the medical field (in treatments like radiation therapy) to environmental science (such as dating of ancient artifacts using carbon-14). It's a field marked by both its potential for generating clean energy and its responsibility in managing nuclear waste and the threats associated with radiation.
Understanding nuclear chemistry is vital for multiple real-world applications ranging from the medical field (in treatments like radiation therapy) to environmental science (such as dating of ancient artifacts using carbon-14). It's a field marked by both its potential for generating clean energy and its responsibility in managing nuclear waste and the threats associated with radiation.
Other exercises in this chapter
Problem 2
Lead-210 is used to prepare eyes for corneal transplants. Its decay product is bismuth-210. Identify the emission from lead-210.
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Write balanced nuclear reactions for the following: (a) Formation of Am-241 through \(\beta\) -emissions. (b) Formation of Kr-81 by K-electron capture. (c) Form
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Write balanced nuclear equations for the following: (a) Formation of Mn-52 by positron emission. (b) Formation of Ac-228 by \(\beta\) -emission. (c) Formation o
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