Chapter 37: Problem 2
A cobalt-60 source has a measured activity of \(12,000 \mathrm{cpm}\). After how long would the observed activity be \(750 \mathrm{cpm} ?\) (The half-life of \(\operatorname{Co}-60\) is \(5.27\) years.)
Short Answer
Expert verified
The observed activity would be 750 cpm after approximately 21.08 years.
Step by step solution
01
Understand the Problem
We need to determine how long it takes for the activity of a cobalt-60 source to decay from 12,000 counts per minute (cpm) to 750 cpm. Given the half-life of Co-60 is 5.27 years, we'll be using the decay formula to solve this.
02
Identify the Decay Formula
Radioactive decay follows the formula \( A(t) = A_0 \times (\frac{1}{2})^{t/T_{1/2}} \), where \( A(t) \) is the final activity, \( A_0 \) is the initial activity, \( t \) is the time elapsed, and \( T_{1/2} \) is the half-life.
03
Plug Values into the Formula
Substitute the given values into the formula: \( 750 = 12000 \times (\frac{1}{2})^{t/5.27} \). Here, \( A(t) = 750 \), \( A_0 = 12,000 \), and \( T_{1/2} = 5.27 \) years.
04
Solve for Time (t)
Rearrange the equation to solve for \( t \): \( (\frac{1}{2})^{t/5.27} = \frac{750}{12000} = \frac{1}{16} \). Now, solve for \( t \) using logarithms: \( t/5.27 = \log_{1/2}(\frac{1}{16}) \).
05
Calculate the Logarithm
Calculate \( \log_{1/2}(\frac{1}{16}) \): Since \( \frac{1}{2}^4 = \frac{1}{16} \), it implies that \( t/5.27 = 4 \).
06
Find the Time (t)
Multiply both sides by 5.27 to find \( t \): \( t = 4 \times 5.27 = 21.08 \).
07
Conclusion
The time it takes for the activity to drop from 12,000 cpm to 750 cpm is approximately 21.08 years.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cobalt-60
Cobalt-60 is a radioactive isotope of cobalt that is commonly used in medical and industrial applications. It is artificially produced in nuclear reactors by irradiating cobalt-59. This isotope is known for
- emitting beta particles and intense gamma rays,
- its use in cancer treatment through radiation therapy,
- and its application in sterilizing medical equipment.
Half-life
The half-life is a key concept in understanding radioactive decay. It refers to the time required for half the quantity of a radioactive substance to decay. For cobalt-60, the half-life is 5.27 years.
- This means that every 5.27 years, the amount of cobalt-60 will reduce to half of its initial value.
- This property makes cobalt-60 particularly useful in applications where long-term stability is needed.
Radioactive Decay Formula
The radioactive decay formula is instrumental in calculating the activity of a radioactive substance over time. The general form of the decay formula is:\[A(t) = A_0 \times \left(\frac{1}{2}\right)^{t/T_{1/2}}\]where:
- \(A(t)\) is the activity after time \(t\),
- \(A_0\) is the initial activity,
- \(T_{1/2}\) is the half-life of the substance.