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 \({ }^{241}\) 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

The number of alpha particles emitted per second by a 5.00 g sample of ${ }^{241}\mathrm{Am}$ can be found in four steps: 1. Calculate the initial number of ${ }^{241}\mathrm{Am}$ nuclei, N₀, in the sample: \(N₀ = \frac{5 \, g}{241 \, g/mol} \times 6.022 \times 10^{23} \, mol^{-1}\) 2. Determine the decay constant `λ` using the half-life of \({ }^{241}\mathrm{Am}\): \(λ = \frac{ln(2)}{T}\), with T = \(433 \, years \times 3.154 \times 10^7 \, s/year\) 3. Calculate the rate of decay `R` using the decay constant and the initial number of nuclei: \(R = λ \times N₀\) 4. The number of alpha particles emitted per second is equal to the rate of decay `R`.

Step-by-step solution

Find the initial number of \({ }^{241}\mathrm{Am}\) nuclei in the sample

First, we need to convert the mass of the sample into the number of \({ }^{241}\mathrm{Am}\) nuclei. The molar mass of \({ }^{241}\mathrm{Am}\) is 241 g/mol. Using Avogadro's number (N_A = \(6.022 \times 10^{23}\) mol⁻¹), we calculate the initial number of \({ }^{241}\mathrm{Am}\) nuclei (N₀) as follows: N₀ = (mass_sample / molar_mass) × N_A N₀ = (5 g / 241 g/mol) × \(6.022 \times 10^{23}\) mol⁻¹

Determine the decay constant `λ`

The half-life (T) of \({ }^{241}\mathrm{Am}\) is given as 433 years. We will convert it to seconds for further calculations. T = 433 years × (3.154 × \(10^7\) s/year) Now, we can calculate the decay constant `λ` using the following formula: \(λ = \frac{ln(2)}{T}\)

Calculate the rate of decay `R`

With the decay constant `λ` and the initial number of \({ }^{241}\mathrm{Am}\) nuclei (N₀), we can compute the rate of decay `R` using the formula: R = λ × N₀

Find the number of alpha particles emitted each second

Knowing that one alpha particle is emitted per decay, we can conclude that the number of alpha particles emitted each second is equal to the rate of decay `R`. Therefore, the solution to this problem is: Number of alpha particles emitted per second = R

Key concepts

Half-Life

The half-life of a radioactive isotope is the amount of time it takes for half of the original number of nuclei to decay. Understanding half-life is crucial when dealing with nuclear decay calculations, as it helps in determining how quickly a radioactive element will become less active over time.

Let's apply this knowledge to our Americium-241 problem from smoke detectors. With its half-life of 433 years, we can deduce that, hypothetically, if we start with a sample size of 1000 nuclei of Americium-241, in 433 years, only about 500 of these nuclei would remain, and the rest would have decayed. This decay rate is an exponential process, meaning it does not occur linearly over time but instead, the rate slows down as the quantity of the substance decreases.

Avogadro's Number

Avogadro's number, famously known as the 'mole' concept in chemistry, is essential for converting between atoms/molecules and grams. In our context, it's defined as the number of atoms found in one mole of any substance and has a value of approximately \(6.022 \times 10^{23}\) entities per mole.

In the case of our Americium-241 sample, we use Avogadro's number to find out how many individual nuclei we have in a 5.00-g sample. By understanding the molar mass of Americium-241 and applying Avogadro's number, we can calculate precisely how many atoms we're dealing with, a fundamental step in finding out the rate at which alpha particles are emitted.

Decay Constant

The decay constant, represented by \(\lambda\), is a probability rate at which a single nucleus will decay per unit time. It is inherently related to, and determined by, the half-life of the radioactive substance. The smaller the half-life, the greater the decay constant, which highlights the inverse relationship between the two.

Our exercise involves deriving the decay constant of Americium-241 from its half-life. The formula \(\lambda = \frac{\ln(2)}{T}\) is used, where \(T\) is the half-life period. Once we know the decay constant, it allows us to predict the rate of decay of our sample at any given moment, leading us to how many nuclei are expected to decay – and hence emit an alpha particle – each second.

Alpha Particle Emission

Alpha particle emission is a common mode of radioactive decay especially in heavy elements, where the nucleus emits an alpha particle (composed of two protons and two neutrons) to form a new element. Each emission event causes the mass number to drop by four and the atomic number by two, leading to the transmutation of the element.

In the scenario with the Americium-241 sample, alpha particle emission is the method of decay. Because each decay event releases one alpha particle, understanding this emission helps us calculate the number of alpha particles released per unit time. Once the decay constant and the initial number of Am nuclei are known, we can precisely calculate the number of alpha particles emitted each second by our sample.