Question

Cobalt-60, which undergoes beta decay, has a half-life of \(5.26 \mathrm{yr}\). (a) How many beta particles arc emitted in \(600 \mathrm{~s}\) by a \(3.75-\mathrm{mg}\) sample of \({ }^{60} \mathrm{Co}\) ? (b) What is the activity of the sample in Bq?

Short answer

(a) In 600 seconds, the 3.75-mg sample of Cobalt-60 will emit approximately \(9.413 \times 10^{16}\) beta particles. (b) The activity of the sample is approximately \(1.574 \times 10^{11}\) Bq.

Step-by-step solution

Calculate the decay constant λ

First, we'll find the decay constant using the given half-life of Cobalt-60 (5.26 years). The equation to be used is: λ = ln(2)/t_1/2 Convert the given half-life into seconds: 5.26 years = 5.26 x 365 x 24 x 60 x 60 = 165900480 s Now, we can find the decay constant: λ = ln(2) / 165900480 ≈ 4.1816 x 10^-9 s^-1.

Find the number of \({ }^{60}\mathrm{Co}\) atoms

To find the number of Cobalt-60 atoms in the 3.75-mg sample, we need to use Avogadro's number. Molar mass of Cobalt-60: 60 g/mol First, convert the mass of the sample from mg to g: 3.75 mg = 0.00375 g Next, find the number of moles of Cobalt-60: n = 0.00375 g / 60 g/mol ≈ 6.25 x 10^-5 mol Then, find the number of Cobalt-60 atoms using Avogadro's number (6.022 x 10^23 atoms/mol): N₀ = 6.25 x 10^-5 mol x 6.022 x 10^23 atoms/mol ≈ 3.764 x 10^19 atoms

Calculate the number of beta particles emitted in 600 s

Now, we'll find the number of beta particles emitted in 600 s using the equation: N(t) = N₀(1 - e ^(-λt)) Plug in the values: N(600) = 3.764 x 10^19 atoms (1 - e^(-4.1816 x 10^-9 s^-1)(600 s)) N(600) = 3.764 x 10^19 atoms (1 - e^-0.002509) N(600) ≈ 3.764 x 10^19 atoms (1 - 0.9975) = 9.413 x 10^16 beta particles

Calculate the activity of the sample in Bq

Finally, we can find the activity of the sample in Bq using the equation: A = λN₀ A = 4.1816 x 10^-9 s^-1 x 3.764 x 10^19 atoms A ≈ 1.574 x 10^11 Bq The activity of the 3.75-mg Cobalt-60 sample is approximately \(1.574 x 10^{11}\) Bq.

Key concepts

Half-life

Half-life is a fundamental concept in radioactive decay describing the time it takes for half of the radioactive atoms in a sample to decay. Given Cobalt-60's half-life of 5.26 years, this means that after 5.26 years, half of the initial amount of Cobalt-60 will have decayed into a different element. This concept helps in understanding how quickly or slowly a radioactive substance transforms over time.

The half-life expression is given by \[ \lambda = \frac{\ln(2)}{t_{1/2}} \]where \(t_{1/2}\) is the half-life, and \(\lambda\) is the decay constant, representing the probability per unit time that a single atom will decay. Half-life is crucial for calculating other aspects of radioactive processes, such as how much radiation a substance will emit at any given time, or how much of a material remains after a certain period.

Beta particles

Beta particles are high-energy, high-speed electrons or positrons emitted by certain types of radioactive nuclei, such as Cobalt-60. When an unstable nucleus emits a beta particle, it undergoes a process called beta decay, whereby a neutron in the nucleus is transformed into a proton. This results in the emission of an electron (beta particle) and an anti-neutrino from the nucleus.

Beta particles are significant in the study of radioactivity because they indicate the instability of the element and the resulting transformation. They hold essential value in medical therapies, like cancer treatment, due to their ability to penetrate biological tissues and deposit energy. Understanding beta decay is useful for calculating the number of particles emitted from a radioactive source over a certain time, which directly ties into determining the potential health risks and other effects.

Cobalt-60

Cobalt-60 is a radioactive isotope of the element Cobalt, represented as \(^ {60} \text{Co}\). It is a prominent source of gamma rays and beta particles. The half-life of Cobalt-60 is 5.26 years, making it an ideal choice for various applications, including radiotherapy in cancer treatment, industrial radiography, and food sterilization.

Its radioactive decay involves the emission of beta particles, which transition the nucleus from Cobalt-60 to Nickel-60. This decay process emits energetic gamma rays, harnessed in multiple technologies. Cobalt-60's stable state after decay and predictable life span make it invaluable both in industry and medicine.

• Medical: Targeted killing of cancer cells.
• Industrial: Non-destructive testing of materials.
• Scientific research: Study of gamma radiation effects.

Understanding Cobalt-60 involves looking at its decay patterns and applications worldwide.

Avogadro's number

Avogadro's number is a pivotal concept in chemistry and physics, defined as \(6.022 \times 10^{23}\), representing the number of atoms, ions, or molecules in one mole of a substance. This constant bridges the macroscopic scale we observe and the microscopic scale at which atoms and molecules exist.

In the context of Cobalt-60, Avogadro's number is used to determine the number of atoms within a given sample, facilitating calculations about its activity and decay. By understanding the relationship between molar mass and the number of atoms, researchers can accurately predict the behaviors of radioactive samples, such as the number of beta particles emitted over time.
The calculation of moles is given by: \[ n = \frac{\text{mass of sample}}{\text{molar mass}} \] and then multiplied by Avogadro's number to find the actual number of atoms. This is crucial for practical applications, including the prediction of decay processes in radioactive samples and the assessment of radioactive activity in experimental settings.