Question

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

The observed activity would be 750 cpm after approximately 21.08 years.

Step-by-step solution

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.

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.

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.

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: \( t/5.27 = \log_{1/2}(\frac{1}{16}) \).

Calculate the Logarithm

Calculate \( \log_{1/2}(\frac{1}{16}) \): Since \( \frac{1}{2}^4 = \frac{1}{16} \), it implies that \( t/5.27 = 4 \).

Find the Time (t)

Multiply both sides by 5.27 to find \( t \): \( t = 4 \times 5.27 = 21.08 \).

Conclusion

The time it takes for the activity to drop from 12,000 cpm to 750 cpm is approximately 21.08 years.

Key concepts

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.

Cobalt-60 decays into stable nickel-60, releasing radiation in the process. Understanding the characteristics of cobalt-60 helps in predicting its behavior under radioactive decay. In safety protocols, its half-life and decay process are critical for handling and disposal, ensuring minimal harm to humans and the environment.

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.

The steady and predictable nature of half-life allows scientists to calculate the rate of decay and the remaining amount of a substance over time. This is crucial, especially when we need to calculate how long it will take for a radioactive source to become safe or to reach a specific level of activity.

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.

In the given exercise, you substitute known values to find the time \(t\) when the activity decreases from 12,000 cpm to 750 cpm. Using logarithms to solve for \(t\), we can determine the elapsed time over which the decay occurs. This formula is a powerful tool in predicting the future activity of radioactive materials, making it essential for many fields, from nuclear physics to environmental science.