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Cobalt-60, which undergoes beta decay, has a half-life of \(5.26\) yr. (a) How many beta particles are emitted in \(180 \mathrm{~s}\) by a \(3.75\) -mg sample of \({ }^{60} \mathrm{Co}\) ? (b) What is the activity of the sample in Bq?

Short Answer

Expert verified
The number of beta particles emitted in 180 seconds is negligible, and the activity of the 3.75 mg Cobalt-60 sample is approximately \(1.57 \times 10^{11}\) Bq.

Step by step solution

01

Finding the decay constant

We have the half-life of Cobalt-60, \(T_{1/2} = 5.26\) years. To find the decay constant, \(\lambda\), we use the equation: \[ \lambda = \frac{\ln 2}{T_{1/2}} \] Converting the half-life to seconds: \[ T_{1/2} = 5.26 \text{ years} \times \frac{365 \text{ days}}{1 \text{ year}} \times \frac{24 \text{ hours}}{1 \text{ day}} \times \frac{60 \text{ minutes}}{1 \text{ hour}} \times \frac{60 \text{ seconds}}{1 \text{ minute}} = 1.66 \times 10^{8} \text{ seconds} \] Now, plugging the value of \(T_{1/2}\) into the decay constant formula: \[ \lambda = \frac{\ln 2}{1.66 \times 10^{8} \text{ s}} \approx 4.18 \times 10^{-9} \text{ s}^{-1} \]
02

Finding the initial number of nuclei

We are given the mass of the sample, \(m = 3.75 \text{ mg}\). To find the initial number of Cobalt-60 nuclei, we will first determine the number of moles in the sample and then multiply that by Avogadro's number, \(N_A\). The atomic mass of Cobalt-60 is 60 g/mol. \[ \text{moles} = \frac{m}{\text{atomic mass}} = \frac{3.75 \times 10^{-3} \text{ g}}{60 \text{ g/mol}} = 6.25 \times 10^{-5} \text{ mol} \] Now, multiplying the number of moles by Avogadro's number (approximately \(6.022 \times 10^{23} \text{ mol}^{-1}\)): \[ N_0 = 6.25 \times 10^{-5} \text{ mol} \times 6.022 \times 10^{23} \text{ mol}^{-1} = 3.76 \times 10^{19} \text{ nuclei} \]
03

Finding the number of beta particles emitted

We have the initial number of nuclei and decay constant. The decay equation can be written as: \[ N(t) = N_0 \cdot e^{-\lambda t} \] where \(N(t)\) is the number of nuclei remaining at time \(t\), \(N_0\) is the initial number of nuclei, \(\lambda\) is the decay constant, and \(t\) is the time period for which we want to find the number of emitted beta particles (180 s in this case). We will find the remaining number of nuclei at time \(t = 180\) seconds and subtract this value from the initial number of nuclei to obtain the number of emitted beta particles: \[ N(180\text{ s}) = 3.76 \times 10^{19} \text{ nuclei} \cdot e^{-4.18 \times 10^{-9} \text{ s}^{-1}\times 180 \text{ s}} \] Calculating \(N(180\text{ s})\): \[ N(180\text{ s}) \approx 3.76 \times 10^{19} \text{ nuclei} \] Since the value is almost the same as the initial number of nuclei, the number of emitted beta particles in 180 seconds is negligible.
04

Finding the activity

The activity, \(A\), of a sample can be calculated using the formula: \[ A = \lambda \cdot N_0 \] where \(\lambda\) is the decay constant and \(N_0\) is the initial number of nuclei. Plugging in the values we found for the decay constant and the initial number of nuclei: \[ A = 4.18 \times 10^{-9} \text{ s}^{-1} \times 3.76 \times 10^{19} \text{ nuclei} \approx 1.57 \times 10^{11} \text{ Bq} \] The activity of the Cobalt-60 sample is approximately \(1.57 \times 10^{11}\) Bq.

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Key Concepts

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

Beta Decay
Beta decay is a type of radioactive decay in which an unstable atomic nucleus transforms and emits a beta particle, which is a high-speed electron or positron. This occurs when a neutron in the nucleus converts to a proton, emitting an electron (beta minus decay), or a proton converts to a neutron, emitting a positron (beta plus decay), along with an antineutrino or neutrino, respectively.

During beta decay, the atomic number of the nucleus changes by one, either increasing (in beta minus decay) or decreasing (in beta plus decay). This process allows the atomic nucleus to move closer to the optimal ratio of protons to neutrons, which increases stability. The emitted beta particles can be detected and measured, providing insight into the properties and behaviors of radioactive materials.
Half-Life
The half-life of a radioactive isotope is the time it takes for half of the isotopes in a given sample to undergo radioactive decay. It is a fundamental concept in nuclear physics that helps us understand how quickly or slowly a radioactive material decays.

Half-life varies widely among different isotopes, from fractions of a second to millions of years. It is a constant property for each radioactive substance, meaning that regardless of the amount of the substance, the proportion of decayed to undecayed nuclei is consistent over each half-life interval. Understanding half-life is essential when calculating how long a sample of radioactive material will remain active, which is crucial in fields like medicine, archaeology, and environmental science.
Decay Constant
The decay constant, often symbolized by \(\lambda\), is a probability factor that defines the rate at which an unstable atomic nucleus decays. It represents the fraction of a sample of radioactive nuclei that decay per unit of time.

In mathematical terms, the decay constant is inversely proportional to the half-life of the isotope—having a higher decay constant means the substance will have a shorter half-life, and thus, it will decay more quickly. This concept is crucial when calculating the age of archaeological finds through radiometric dating or planning the health and safety precautions necessary for handling radioactive materials.
Activity (Becquerel)
Activity is a measure of the rate of radioactive decay in a sample, and it is quantified in the unit of Becquerel (Bq), where one Bq corresponds to one decay event per second. High activity means a large number of atomic nuclei are decaying and emitting radiation each second.

By measuring activity, it is possible to estimate the radiation exposure risk and the effectiveness of radioactive materials used in medical treatments. For instance, in the provided exercise, calculating the activity of a Cobalt-60 sample involved multiplying its decay constant by the initial number of nuclei, resulting in an activity expressed in becquerels. Understanding activity levels is essential for applications involving nuclear medicine, radiation therapy, and ensuring safety standards in environments exposed to radioactive substances.

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Most popular questions from this chapter

Based on the following atomic mass values \(-{ }^{1} \mathrm{H}\), \(1.00782 \mathrm{amu} ;{ }^{2} \mathrm{H}, 2.01410 \mathrm{amu} ;{ }^{3} \mathrm{H}, 3.01605 \mathrm{amu} ;{ }^{3} \mathrm{He}\), \(3.01603\) amu; \({ }^{4} \mathrm{He}, 4.00260\) amu-and the mass of the neutron given in the text, calculate the energy released per mole in each of the following nuclear reactions, all of which are possibilities for a controlled fusion process: (a) \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) (b) \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{3} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) (c) \({ }_{1}^{2} \mathrm{H}+{ }_{2}^{3} \mathrm{He} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{1} \mathrm{H}\)

A 25.0-mLsample of \(0.050 \mathrm{M}\) barium nitrate solution was mixed with \(25.0 \mathrm{~mL}\) of \(0.050 \mathrm{M}\) sodium sulfate solution labeled with radioactive sulfur-35. The activity of the initial sodium sulfate solution was \(1.22 \times 10^{6} \mathrm{~Bq} / \mathrm{mL}\). After the resultant precipitate was removed by filtration, the remaining filtrate was found to have an activity of \(250 \mathrm{~Bq} / \mathrm{mL}\). (a) Write a balanced chemical equation for the reaction that occurred. (b) Calculate the \(K_{s p}\) for the precipitate under the conditions of the experiment.

A free neutron is unstable and decays into a proton with a half-life of \(10.4\) min. (a) What other particle forms? (b) Why don't neutrons in atomic nuclei decay at the same rate?

Cobalt-60 is a strong gamma emitter that has a half-life of \(5.26 \mathrm{yr}\). The cobalt-60 in a radiotherapy unit must be replaced when its radioactivity falls to \(75 \%\) of the original sample. (a) If an original sample was purchased in June 2006, when will it be necessary to replace the cobalt-60? (b) How can you store cobalt-60 so that it is safe to handle?

Complete and balance the following nuclear equations by supplying the missing particle: (a) \({ }^{25}{ }_{8}^{2} \mathrm{Cf}+{ }_{5}^{10} \mathrm{~B} \longrightarrow 3_{0}^{1} \mathrm{n}+?\) (b) \({ }_{1}^{2} \mathrm{H}+{ }_{2}^{3} \mathrm{He} \longrightarrow{ }_{2}^{4} \mathrm{He}+?\)

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