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Chapter 21: Problem 35

Cobalt- 60 is a strong gamma emitter that has a half-life of 5.26 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 2010 , when will it be necessary to replace the cobalt- \(60 ?\)

Short Answer

Expert verified
The cobalt-60 sample will need to be replaced in February 2012.

Step by step solution

01

Understand the problem

In this problem, we are given the half-life of cobalt-60 (5.26 years) and are asked to determine when the radioactivity falls to 75% of its original amount.
02

Write the formula for radioactive decay

To solve this problem, we can use the half-life formula for radioactive decay: \[N_t = N_0 \cdot (1/2)^{\frac{t}{t_{1/2}}}\] Where \(N_t\) is the radioactivity at time t, \(N_0\) is the initial radioactivity, t is the time, and \(t_{1/2}\) is the half-life.
03

Set up the equation

We are given that the sample's radioactivity must be 75% of its original amount, so we can set up the equation: \[0.75N_0 = N_0 \cdot (1/2)^{\frac{t}{5.26}}\]
04

Solve for time t

Now, we can solve for the time t when the radioactivity falls to 75%: \[\frac{0.75N_0}{N_0} = (1/2)^{\frac{t}{5.26}}\] \[0.75 = (1/2)^{\frac{t}{5.26}}\] To solve for t, we can use the logarithm: \[t = 5.26 \cdot \log_{1/2}(0.75)\] Use a calculator to find the value of t: \[t \approx 1.66 \text{ years}\]
05

Find the replacement date

The cobalt-60 sample was purchased in June 2010, so we need to add the time until the radioactivity falls to 75% to this date: June 2010 + 1.66 years ≈ June 2010 + 1 year and 8 months ≈ February 2012 So, it will be necessary to replace the cobalt-60 sample in February 2012.

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