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Chapter 21: Problem 37

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. If an original sample was purchased in June \(2016,\) when will it be necessary to replace the cobalt- \(60 ?\)

Short Answer

Expert verified
The Cobalt-60 in the radiotherapy unit will need to be replaced in March 2018, as it takes approximately 1.84 years for its radioactivity to decrease to 75% of its original value.

Step by step solution

01

Identify the formula for radioactive decay.

The formula for radioactive decay is given by \[A_t = A_0 \cdot (1/2)^{\frac{t}{t_{1/2}}}\] Where, \(A_t\) = Activity at time t, \(A_0\) = Initial activity, \(t\) = Time (in years), \(t_{1/2}\) = Half-life of the substance (in years).
02

Set up the equation using the given information.

We know that we have to replace Cobalt-60 when its radioactivity falls to \(75\%\) of the original sample. So, \(A_t = 0.75A_0\). The half-life of Cobalt-60 is given as 5.26 years. We need to find the time t when the activity reduces to 75%. So, the equation becomes: \[0.75A_0 = A_0 \cdot (1/2)^{\frac{t}{5.26}}\]
03

Solve for the time t.

As we are interested in finding the time t, we can eliminate the term \(A_0\) from the equation: \[\frac{0.75A_0}{A_0} = (1/2)^{\frac{t}{5.26}}\] Now, take the natural logarithm of both sides of the equation: \[\ln{0.75} = \ln{(1/2)^{\frac{t}{5.26}}}\] Next, use the logarithm power rule to move the exponent to the front: \[\ln{0.75} = \frac{t}{5.26} \ln{\frac{1}{2}}\] Now, solve for t by isolating t on one side of the equation: \[t = 5.26 \cdot \frac{\ln{0.75}}{\ln{\frac{1}{2}}}\] Calculate t: \[t \approx 1.84\]
04

Determine the replacement date of Cobalt-60.

The original sample was purchased in June 2016. It will take about 1.84 years for the radioactivity to decrease to 75% of its original value. Thus, we need to replace the Cobalt-60 after 1.84 years from June 2016. Let's convert 1.84 years to months: \[1.84 \mathrm{years} \times \frac{12 \mathrm{months}}{1 \mathrm{year}} \approx 22.08 \mathrm{months}\] Round this value to the nearest whole number, which is 22 months. Count 22 months from June 2016: June (1), July (2), August (3), ... , March (22) So, it will be necessary to replace the Cobalt-60 in March 2018.

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Most popular questions from this chapter

Chlorine has two stable nuclides, \({ }^{35} \mathrm{Cl}\) and ${ }^{37} \mathrm{Cl}\(. In contrast, \){ }^{36} \mathrm{Cl}$ is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of \({ }^{36} \mathrm{Cl}\) ? (b) Based on the empirical rules about nuclear stability, explain why the nucleus of \({ }^{36} \mathrm{Cl}\) is less stable than either ${ }^{35} \mathrm{Cl}\( or \){ }^{37} \mathrm{Cl}$.

In 2002 , a team of scientists from Russia and the United States reported the creation of the first atom of element 118 , which is named oganesson, and whose symbol is Og. The synthesis involved the collision of californium- 249 atoms with accelerated ions of an atom which we will denote X. In the synthesis, an oganesson-294 is formed together with three neutrons. $$ { }_{98}^{249} \mathrm{Cf}+\mathrm{X} \longrightarrow{ }_{118}^{294} \mathrm{Og}+3{ }_{0}^{1} \mathrm{n} $$ (a) What are the identities of isotopes X? (b) Isotope \(X\) is unusual in that it is very long-lived (its half-life is on the order of \(10^{19} \mathrm{yr}\) ) in spite of having an unfavorable neutron-to-proton ratio (Figure 21.1). Can you propose a reason for its unusual stability? (c) Oganesson-294 decays into livermorium-290 by alpha decay. Write a balanced equation for this.

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