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Chapter 21: Problem 38

Cobalt- 60 , which undergoes beta decay, has a half-life of 5.26 yr. (a) How many beta particles are 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 \(\mathrm{Bq}\) ?

Short Answer

Expert verified
The number of Cobalt-60 nuclei in the $3.75 \mathrm{ ~mg}$ sample can be calculated using the formula \(N = \frac{m \cdot N_A}{M}\), giving us \(N \approx 3.77 \times 10^{19}\) nuclei. The decay constant \(\lambda\) is found using the half-life formula, resulting in \(\lambda \approx 1.32 \times 10^{-11} \mathrm{~s}^{-1}\). The number of beta particles emitted in $600 \mathrm{~s}$ can be calculated as \(N_{\text{decay}} = N \cdot (1 - e^{-\lambda t})\), giving us \(N_{\text{decay}} \approx 3.14 \times 10^{9}\) beta particles. The activity of the sample in Bq can be calculated as \(A = \lambda \cdot N\), resulting in \(A \approx 4.98 \times 10^{8} \mathrm{~Bq}\).

Step by step solution

01

Calculate the number of Cobalt-60 nuclei in the sample

To find the total number of Cobalt-60 nuclei in the sample, we will use the formula: \[N = \frac{m \cdot N_A}{M}\] Where N is the number of nuclei, m is the mass of the sample, \(N_A\) is Avogadro's constant \((6.022 \times 10^{23} \ \text{mol}^{-1})\), and M is the molar mass of Cobalt-60 which is 59.9338 u. First, we need to convert the mass of the sample to grams: \[3.75 \ \text{mg} = 3.75 \times 10^{-3} \ \text{g}\] Now we can plug in the values and calculate the number of nuclei: \[N = \frac{(3.75 \times 10^{-3} \ \text{g}) \times (6.022 \times 10^{23} \ \text{mol}^{-1}) }{59.9338 \ \text{g/mol}}\]
02

Calculate the decay constant

To find the decay constant, we will use the half-life formula: \[\lambda = \frac{\ln{2}}{t_{1/2}}\] Where \(\lambda\) is the decay constant and \(t_{1/2}\) is the half-life of the substance. We are given the half-life of Cobalt-60 as 5.26 yr. Plugging in the values, we get: \[\lambda = \frac{\ln{2}}{5.26 \ \text{yr}}\]
03

Find the number of beta particles emitted in 600 s

Now that we have the decay constant, we can find the number of beta particles emitted during a specific time interval using the formula: \[N_{\text{decay}} = N \cdot (1 - e^{-\lambda t})\] We are given a time interval of 600 s. Since we have calculated \(\lambda\) in years, we should convert the time to years: \[600 \ \text{s} = \frac{600}{60 \times 60 \times 24 \times 365.25} \ \text{yr}\] Now we can calculate the number of beta particles emitted: \[N_{\text{decay}} = N \cdot (1 - e^{-\lambda t})\]
04

Calculate the activity in Bq

The activity of the sample is defined as the rate of decay, which we can find using the decay constant and the number of nuclei in the sample: \[A = \lambda \cdot N\] The activity should be expressed in Bq which corresponds to decays per second (s^-1). Since we have found \(\lambda\) in year^-1, we should convert it to s^-1: \[\lambda = \frac{\lambda}{(60 \times 60 \times 24 \times 365.25) \ s^{-1}}\] Now we can calculate the activity: \[A = \lambda \cdot N\]

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Most popular questions from this chapter

Which type or types of nuclear reactors have these characteristics? (a) Does not use a secondary coolant (b) Creates more fissionable material than it consumes (c) Uses a gas, such as He or \(\mathrm{CO}_{2}\), as the primary coolant

Charcoal samples from Stonehenge in England were burned in \(\mathrm{O}_{2},\) and the resultant \(\mathrm{CO}_{2}\) gas bubbled into a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) (limewater), resulting in the precipitation of \(\mathrm{CaCO}_{3} .\) The \(\mathrm{CaCO}_{3}\) was removed by filtration and dried. A \(788-\mathrm{mg}\) sample of the \(\mathrm{CaCO}_{3}\) had a radioactivity of \(1.5 \times 10^{-2} \mathrm{~Bq}\) due to carbon-14. By comparison, living organisms undergo 15.3 disintegrations per minute per gram of carbon. Using the half-life of carbon- \(14,5715 \mathrm{yr},\) calculate the age of the charcoal sample.

Give the symbol for (a) a proton, (b) a beta particle, (c) a positron.

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Indicate the number of protons and neutrons in the following nuclei: (a) \({ }_{22}^{55} \mathrm{Mn},(\mathbf{b}){ }^{201} \mathrm{Hg},(\mathbf{c})\) potassium- \(39 .\)
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