Problem 26 - Chemistry An Atoms First Approach

Question

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 -g sample of \(^{241} \mathrm{Am} ?\)

Step-by-Step Solution

Answer

The number of alpha particles emitted each second by a 5.00-g sample of $^{241}\mathrm{Am}$ can be calculated using the decay law: \(R = \frac{5.00}{241} × 6.022 \times 10^{23} × \frac{ln(2)}{433 \times (3.1536\times10^7)}\). Upon calculating this expression, we will get the number of alpha particles emitted per second.

Step 1: Calculate the Moles of Americium-241

First, we need to find the number of moles in the 5g sample. We can do this using the molar mass of Americium-241 (Am) which is 241 g/mol. Moles of Am = \(\frac{\text{mass}}{\text{molar mass}}\) Moles of Am = \(\frac{5.00}{241}\)

Step 2: Calculate the Number of Am-241 Atoms

Next, we will calculate the number of Americium-241 atoms in the sample. To do this, we need to multiply the number of moles by Avogadro's number (NA ≈ \(6.022 \times 10^{23}\) atoms/mol). Number of Am-241 atoms = Moles of Am × NA Number of Am-241 atoms = \(\frac{5.00}{241} × 6.022 \times 10^{23}\)

Step 3: Calculate the Decay Constant

We are given the half-life of Am-241 is 433 years. We can use this information to find the decay constant (λ). The decay constant and half-life are related by the following formula: Half-life (T₁/₂) = \(\frac{ln(2)}{λ}\) Now we can find λ: λ = \(\frac{ln(2)}{433 \text{ years}}\) Since we need the decay constant in seconds, we need to convert years to seconds. There are \(3.1536\times10^7\) seconds in a year. λ = \(\frac{ln(2)}{433 \times (3.1536\times10^7) \text{ seconds}}\)

Step 4: Calculate the Number of Decays per Second

Now we can use the decay law to find the number of atoms decaying each second, which is equal to the number of alpha particles emitted. Decays per second (R) = Number of Am-241 atoms × λ R = \(\frac{5.00}{241} × 6.022 \times 10^{23} × \frac{ln(2)}{433 \times (3.1536\times10^7)}\)

Step 5: Find the Number of Alpha Particles Emitted per Second

Since each decay emits one alpha particle, the number of alpha particles emitted each second will be equal to the number of decays per second calculated in step 4. Number of alpha particles per second = R By solving for R, we can find out how many alpha particles are emitted each second by a 5.00-g sample of Am-241.

Key Concepts

Radioactive Decay

Radioactive decay is a natural process through which an unstable atomic nucleus loses energy by emitting radiation. There are several types of radioactive decay, including alpha, beta, and gamma decay. The decay occurs because the original nucleus, referred to as the parent, transforms into a different nucleus, known as the daughter, reaching a more stable state.

In our exercise, Americium-241 (\( ^{241}\text{Am} \) undergoes radioactive decay by emitting alpha particles. The rate of decay is exponential, and can be described mathematically by the decay law, which states that the activity, or the rate at which the atoms decay, is proportional to the number of unstable nuclei in the sample. This intrinsic property of a nucleus to decay is quantified through its decay constant, symbolized by \(\lambda \).

Half-life of Isotopes

The half-life of an isotope is the time it takes for half of a sample of radioactive material to decay. This concept is fundamental to nuclear chemistry, as it helps predict the longevity and activity of radioactive samples.

In the case of Americium-241, with a half-life of 433 years, after this period, only half of the original nuclei present in any given sample will remain; the rest will have decayed. To calculate specific decay rates or predict the amount of material remaining after a certain time, we utilize the relationship between half-life, \( T_{\frac{1}{2}} \), and the decay constant, \( \lambda \), given by the formula \( T_{\frac{1}{2}} = \frac{\ln(2)}{\lambda} \). The calculation involves natural logarithms and the conversion of time units to match the context of the problem (e.g., seconds, years).

Alpha Particle Emissions

Alpha particle emissions are a type of radioactive decay where the nucleus releases an alpha particle, which consists of two protons and two neutrons, effectively behaving like a helium-4 nucleus. Due to their relatively large mass and charge, alpha particles interact strongly with matter, losing energy quickly and usually only traveling a short distance. This makes them relatively safe when it comes to external exposure, but dangerous if ingested or inhaled.

When Americium-241 decays by emitting an alpha particle, its atomic number decreases by two and its mass number decreases by four, transforming into Neptunium-237. As stated in the exercise, we aim to calculate the number of alpha particles emitted per second, connecting the physical decay process with a quantifiable measure of radioactivity.

Avogadro's Number

Avogadro's number, approximately \(6.022 \times 10^{23} \), is fundamental in chemistry. It is the number of atoms, ions, or molecules in one mole of any substance. Avogadro's number facilitates the conversion between microscopic atomic scale events to macroscopic amounts that we can measure and observe.

In nuclear decay calculations, Avogadro’s number allows us to move from the molar quantity of a substance to the actual number of atoms present, which we require to calculate decay events in a given sample. The relationship between moles, Avogadro's number, and the number of entities is critical for all subsequent calculations, allowing us to determine the amount of radioactive decays, and thus, for example, the number of alpha particles emitted per second.