Question

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

Short answer

In 600 seconds, a 3.75-mg sample of Cobalt-60 emits approximately \(N_\beta = \frac{3.75}{59.93} \times 6.022 \times 10^{23} - \frac{3.75}{59.93} \times 6.022 \times 10^{23}\mathrm{e}^{-\frac{ln(2) \times 600}{5.26 \times 365 \times 24 \times 3600}}\) beta particles. The activity of the sample is \(A = \frac{ln(2)}{5.26 × 365 × 24 × 3600} \times \frac{3.75}{59.93} \times 6.022 \times 10^{23}\mathrm{e}^{-\frac{ln(2) \times 600}{5.26 \times 365 \times 24 \times 3600}}\) Bq.

Step-by-step solution

Find the decay constant λ

Since we know the half-life of Cobalt-60, we can find the decay constant λ using the following formula: \(λ = \frac{ln(2)}{t_{1/2}}\) where \(t_{1/2}\) is the half-life of the substance. For Cobalt-60, the half-life, \(t_{1/2}\) is 5.26 years. To convert this into seconds, we have: \(t_{1/2} = 5.26 × and2$$\Year \times 365 \mathrm{d\ year}^{-1} \times \24 \mathrm{h \ day}^{-1} \times 3600 \mathrm{s \ h}^{-1}\) Plugging this value into the decay constant formula, we get: \(λ = \frac{ln(2)}{5.26 × 365 × 24 × 3600}\)

Calculate the number of atoms in the sample.

We are given a 3.75-mg sample of Cobalt-60. To find the number of atoms in the sample, we need to convert this value to moles, and then multiply by Avogadro's number: Number of moles = \(\frac{\text{mass of the sample}}{\text{molar mass of Cobalt-60}}\) Number of atoms = moles * Avogadro's number The molar mass of Cobalt-60 is 59.93 g/mol. So, the number of moles in the sample is: 3.75 \(\times \mathrm{10}\)^{-3} and$$\= \frac{3.75}{59.93}\)) Now, we can find the number of atoms in the sample by multiplying the number of moles by Avogadro's number (6.022 × 10^23 atoms/mol): Number of atoms = \(\frac{3.75}{59.93} \times 6.022 \times 10^{23}\)

Find the number of beta particles emitted in 600 seconds.

The radioactive decay formula is given by: \(N_t = N_0\mathrm{e}^{-λt}\) where \(N_t\) is the number of atoms left after time \(t\), \(N_0\) is the initial number of atoms, and λ is the decay constant. We want to find the number of atoms that decayed in 600 seconds, or the number of beta particles emitted, which can be found as the difference between the initial number of atoms (\(N_0\)) and the number of atoms left after 600 seconds (\(N_t\)): \(N_\beta = N_0 - N_t = N_0 - N_0\mathrm{e}^{-λt}\) We already have the values for the decay constant λ and the initial number of atoms \(N_0\). We can plug in the values and calculate the number of beta particles emitted in the 600 seconds: \(N_\beta = \frac{3.75}{59.93} \times 6.022 \times 10^{23} - \frac{3.75}{59.93} \times 6.022 \times 10^{23}\mathrm{e}^{-\frac{ln(2) \times 600}{5.26 \times 365 \times 24 \times 3600}}\)

Calculate the activity of the sample in Bq.

The activity of a radioactive sample is defined as the number of decays per unit time. Since the decay rate is equal to the decay constant λ multiplied by the number of atoms at any time t, we can find the activity by plugging in the values for \(N_t\) and λ: Activity (A) = λ × \(N_t\) Using the radioactive decay formula and the values we calculated earlier, we can find the activity: \(A = \frac{ln(2)}{5.26 × 365 × 24 × 3600} \times \frac{3.75}{59.93} \times 6.022 \times 10^{23}\mathrm{e}^{-\frac{ln(2) \times 600}{5.26 \times 365 \times 24 \times 3600}}\) Now we have calculated the activity in the units of decays per second, which is the same as the unit Becquerel (Bq).

Key concepts

Understanding Half-Life in Radioactive Decay

The concept of half-life is pivotal in the field of nuclear physics and plays a crucial role in understanding radioactive decay. A half-life is the duration required for half the atoms in a radioactive sample to decay. This time period is a fixed characteristic for each radioactive isotope, such as Cobalt-60's 5.26 years. To calculate the half-life in practical situations, one must often convert years to seconds to match standard scientific units.

Determining the half-life is not just academic; it has real-world applications, such as dating archaeological artifacts, medical diagnoses through radiopharmaceuticals, and controlling the dosage in radiation therapy. By knowing the half-life, scientists and engineers can predict how long a radioactive material will remain active or hazardous, plan the safe storage of nuclear waste, and track environmental contamination.

The Significance of the Radioactive Decay Constant

The decay constant, denoted by the symbol \(λ\), is the probability per unit time that an atom will decay. It is intimately linked to the concept of half-life, as the two can be used to calculate each other. The formula to find the decay constant using the half-life is expressed as \(λ = \frac{\ln(2)}{t_{1/2}}\), where \(\ln(2)\) represents the natural logarithm of 2.

The importance of the decay constant lies in its use in calculating both the activity of a radioactive sample and predicting future behavior of a sample. For example, in the case of Cobalt-60, once \(λ\) is known, number crunching through various formulas lets us calculate how intense the radioactive source is and how many atoms decay over a specific time, providing critical information for safety and scientific measurements.

Measuring the Activity of a Radioactive Sample

The activity of a radioactive sample provides a measure of the rate at which the atoms within the sample are decaying. It's conventionally expressed in becquerels (Bq), where one Bq corresponds to one decay per second. Calculating the activity involves using the decay constant and the current number of undecayed atoms in the sample. The fundamental equation for activity is \(A = λ × N_t\), with \(N_t\) being the number of atoms at time \(t\).

The practical applications for understanding the activity of a sample are numerous. For instance, in medicine, it allows medical professionals to determine the correct dosage for patients undergoing treatment. In environmental science, it helps in monitoring and managing exposure to radiation. Also, in nuclear power generation, it ensures the reaction rates are kept within safe operating limits.