Americium-241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-life of \({ }^{241} \mathrm{Am}\) is 433 years, and it decays by emitting \(\alpha\) particles. How many \(\alpha\) particles are emitted each second by a \(5.00-\mathrm{g}\) sample of \({ }^{241} \mathrm{Am} ?\)

Short Answer

Expert verified
A 5.00-g sample of Americium-241 emits \(6.33 \times 10^{10}\) alpha particles each second.

Step by step solution

01

Find the number of moles of Americium-241 in the sample

To find the number of moles in the 5.00-g sample of Americium-241, we use the formula: moles = mass / molar_mass The molar mass of Americium-241 is 241 g/mol. Therefore, the number of moles of Americium-241 in the 5.00-g sample is: moles = 5.00 g / 241 g/mol = 0.0207 mol
02

Calculate the number of atoms in the sample

To calculate the number of atoms in the 5.00-g sample, we use Avogadro's number (6.022 x 10^23 atoms/mol): number_of_atoms = moles x Avogadro's_number = 0.0207 mol × 6.022 × 10^23 atoms/mol = 1.25 × 10^22 atoms
03

Calculate the decay constant

The decay constant, λ, can be found using the formula: λ = ln(2) / half-life = ln(2) / 433 years We need to convert the half-life to seconds (1 year = 3.1536 × 10^7 seconds): half-life = 433 years × 3.1536 × 10^7 seconds/year = 1.368 × 10^10 seconds Now, we can find the decay constant: λ = ln(2) / (1.368 × 10^10 seconds) = 5.06 × 10^{-12} s^{-1}
04

Calculate the decay rate (activity) of the sample

The activity (A) of a radioactive sample is given by the formula: A = λ × number_of_atoms Substitute the values found in Steps 2 and 3: A = (5.06 × 10^{-12} s^{-1}) × (1.25 × 10^22 atoms) = 6.33 × 10^10 alpha particles/s
05

Determine the number of alpha particles emitted each second

Since one alpha particle is emitted for each decay event, the number of alpha particles emitted each second by the 5.00-g Americium-241 sample is the same as the activity calculated in Step 4: number_of_alpha_particles = 6.33 × 10^10 alpha particles/s Thus, a 5.00-g sample of Americium-241 emits 6.33 × 10^10 alpha particles each second.

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Key Concepts

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

Half-life Calculation
The concept of half-life is central to the study of radioactive decay. It’s the amount of time required for half of a radioactive substance to decay. To put it simply, if you start with a certain amount of a radioactive isotope, after one half-life, you'll have half the original amount left. In the case of Americium-241, the half-life is quite extensive, amounting to 433 years.

When we talk about half-life, we are often interested in the decay constant (\r). This constant is a probability rate that helps us understand how quickly an isotope decays. The decay constant is linked to the half-life by the natural logarithm of 2, as seen in the formula: \r = \(\frac{ln(2)}{\text{half-life}}\).

To make practical calculations, like finding the decay rate of a sample, it's necessary to express half-life in seconds because rates of decay are typically given per second. Once the decay constant is determined, we can calculate the activity or rate at which the substance is decaying. This rate is essentially the number of disintegrations per second, allowing us to understand the number of alpha particles emitted from a radioactive material in a given timeframe.
Alpha Particle Emission
Alpha particle emission is one kind of radioactive decay, where an unstable atom ejects an alpha particle to become more stable. An alpha particle is made up of two protons and two neutrons, essentially a helium nucleus. When Americium-241 decays, it emits these alpha particles.

This form of decay decreases the atomic number of the original element by two and the mass number by four, leading to the formation of a new element. Alpha particles are relatively heavy and carry a double positive charge, which makes them highly ionizing but not very penetrating. They can be stopped by just a sheet of paper or a few centimeters of air.

In practical applications like smoke detectors, the ionizing capability of alpha particles is harnessed to detect smoke. These particles ionize air molecules, and this ionized air helps to carry a current between two electrodes. If smoke enters the detector, it disrupts the flow of ions and diminishes the current, triggering the alarm. This is directly tied to the amount of alpha emission from materials like Americium-241.
Americium-241
Americium-241 is an artificial radioactive isotope that's mostly used in smoke detectors, as it's a strong alpha emitter. The name 'Americium' comes from the element being discovered in America. With a half-life of 433 years, it poses a long-lasting radioactive source for such devices. Over time, Americium-241 decays into Neptunium-237 while emitting alpha particles.

The relatively long half-life means that a small amount of Americium-241 can last for years in a smoke detector without needing to be replaced. However, though its decay by alpha emission is slow, it is still significant in the context of the life of a smoke detector, and its activity level can eventually drop to a point where the detector becomes ineffective. Proper disposal of Americium-241 is essential due to its long-term radioactivity.
Avogadro's Number
Avogadro's number is a basic chemistry concept and is pivotal in the field of chemistry and physics. It's defined as the number of atoms, ions, or molecules in one mole of substance and is equal to approximately \(6.022 \times 10^{23}\). This constant allows chemists and physicists to calculate the number of particles in a given sample when combined with the mole concept.

For instance, in our textbook exercise, Avogadro's number enables us to convert moles of Americium-241 into the actual number of atoms present in that amount. Once you have the number of atoms and the decay constant, as we previously discussed, you can determine the activity of the sample. Knowing the activity and the nature of the decay – in this case, that each decay corresponds to the emission of one alpha particle – you can directly calculate the number of emissions (or disintegrations) per second, giving a clear understanding of the sample's behavior over time.

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Most popular questions from this chapter

A positron and an electron can annihilate each other on colliding, producing energy as photons: $$ { }_{-1}^{0} \mathrm{e}+{ }_{+1}^{0} \mathrm{e} \longrightarrow 2{ }^{0}{ }_{0}^{0} \gamma $$ Assuming that both \(\gamma\) rays have the same energy, calculate the wavelength of the electromagnetic radiation produced.

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Many transuranium elements, such as plutonium-232, have very short half-lives. (For \({ }^{232} \mathrm{Pu}\), the half-life is 36 minutes.) However, some, like protactinium-231 (half-life \(=3.34 \times\) \(10^{4}\) years), have relatively long half-lives. Use the masses given in the following table to calculate the change in energy when 1 mole of \({ }^{232}\) Pu nuclei and 1 mole of \({ }^{231}\) Pa nuclei are each formed from their respective number of protons and neutrons.

During the research that led to production of the two atomic bombs used against Japan in World War II, different mechanisms for obtaining a supercritical mass of fissionable material were investigated. In one type of bomb, a "gun" shot one piece of fissionable material into a cavity containing another piece of fissionable material. In the second type of bomb, the fissionable material was surrounded with a high explosive that, when detonated, compressed the fissionable material into a smaller volume. Discuss what is meant by critical mass, and explain why the ability to achieve a critical mass is essential to sustaining a nuclear reaction.

Fresh rainwater or surface water contains enough tritium \(\left({ }_{1}^{3} \mathrm{H}\right)\) to show \(5.5\) decay events per minute per \(100 . \mathrm{g}\) water. Tritium has a half-life of \(12.3\) years. You are asked to check a vintage wine that is claimed to have been produced in \(1946 .\) How many decay events per minute should you expect to observe in \(100 . \mathrm{g}\) of that wine?

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