Question

Cobalt-60 (Co-60) has a half-life of approximately 5.3 years. a) Write an exponential function to model this situation. b) What fraction of a sample of Co-60 will remain after 26.5 years? c) How long will it take for a sample of \(\mathrm{Co}-60\) to decay to \(\frac{1}{512}\) of its original mass?

Short answer

a) \[ N(t) = N_0 \times \frac{1}{2}^{\frac{t}{5.3}} \]. b) \frac{1}{32}\ of the sample remains after 26.5 years. c) It takes 47.7 years for the sample to decay to \frac{1}{512}\ of its original mass.

Step-by-step solution

Understanding the half-life formula

The formula for the exponential decay of a substance is given by \[ N(t) = N_0 \times \frac{1}{2}^{\frac{t}{T}} \]where:- \(N(t)\) is the amount remaining after time \(t\),- \(N_0\) is the original amount,- \(T\) is the half-life,- \(t\) is the time elapsed.

Writing the exponential function (Part a)

Given that the half-life of Co-60 is 5.3 years, substitute \(T = 5.3\) into the formula. The exponential function modeling the situation is:\[ N(t) = N_0 \times \frac{1}{2}^{\frac{t}{5.3}} \]

Calculate fraction remaining after 26.5 years (Part b)

To find the fraction remaining after 26.5 years, substitute \(t = 26.5\) into the function:\[ N(26.5) = N_0 \times \frac{1}{2}^{\frac{26.5}{5.3}} \]Simplify the exponent:\[ \frac{26.5}{5.3} = 5 \]Thus:\[ N(26.5) = N_0 \times \frac{1}{2}^5 \]\[ N(26.5) = N_0 \times \frac{1}{32} \]Therefore, the fraction remaining is \( \frac{1}{32} \).

Determine the time to decay to \( \frac{1}{512} \) of original mass (Part c)

Given that \( N(t) = \frac{N_0}{512} \), you need to solve for \( t \) in the equation:\[ \frac{N_0}{512} = N_0 \times \frac{1}{2}^{\frac{t}{5.3}} \]Divide both sides by \( N_0 \):\[ \frac{1}{512} = \frac{1}{2}^{\frac{t}{5.3}} \]Recognize that \( \frac{1}{512} = \left(\frac{1}{2}\right)^9 \):\[ \left(\frac{1}{2}\right)^9 = \left(\frac{1}{2}\right)^{\frac{t}{5.3}} \]Since the bases are equal, equate the exponents:\[ 9 = \frac{t}{5.3} \]Solve for \( t \):\[ t = 9 \times 5.3 \]\[ t = 47.7 \] years.

Key concepts

half-life

Half-life is a fundamental concept in understanding how substances decay over time, especially radioactive materials. It is defined as the time it takes for half of the original amount of a substance to decay or be reduced to half its initial value. In simpler terms, if you start with a certain quantity of a substance, after one half-life, only half of it remains. This process continues, with half of the remaining substance decaying each half-life period. The formula to express this decay is given by:
\( N(t) = N_0 \times \frac{1}{2}^{\frac{t}{T}} \)
where:

• \( N(t) \) is the amount remaining after time \( t \)


• \( N_0 \) is the original amount


• \( T \) is the half-life


• \( t \) is the time elapsed

Half-life is crucial in fields like chemistry, physics, biology, and environmental science because it helps predict how long a substance will remain active or hazardous.

exponential functions

Exponential functions are widely used in various scientific disciplines to model growth or decay processes. These functions grow or decay at rates proportional to their current value, leading to rapid increases or decreases. The general format for an exponential function is:
\( N(t) = N_0 \times e^{kt} \)
However, for situations involving half-lives, we use the formula:
\( N(t) = N_0 \times \frac{1}{2}^{\frac{t}{T}} \)
This version is more intuitive for decay processes. Exponential functions are essential for modeling radioactive decay because they precisely describe how quantities diminish over time, making them invaluable in predicting future amounts of remaining substance.

radioactive decay

Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. This decay process transforms the nucleus into a more stable state, frequently changing the element itself. Radioactive decay can occur in several forms, including alpha decay, beta decay, and gamma decay.
A key aspect of radioactive decay is that it follows an exponential decay pattern, described by the half-life of a substance. The half-life provides a measure of the time it takes for half of the atoms in a radioactive sample to decay. Over multiple half-lives, the quantity of radioactive atoms decreases significantly, eventually approaching zero but never quite reaching it.
Understanding radioactive decay is crucial in fields like nuclear physics, medicine (radiation therapy), and environmental science.

Cobalt-60 decay

Cobalt-60 (Co-60) is a radioactive isotope of cobalt, commonly used in medical treatments and industrial applications. Its decay is notable because its half-life is approximately 5.3 years, making it ideal for certain applications due to its predictable decay rate.
The decay process of Co-60 can be modeled using the exponential decay formula:
\( N(t) = N_0 \times \frac{1}{2}^{\frac{t}{5.3}} \)
For example, if you start with a sample of Co-60, after 26.5 years (which is 5 half-lives), only \( \frac{1}{32} \) of the original sample will remain. Similarly, to determine how long it takes for Co-60 to decay to \( \frac{1}{512} \) of its original mass, you can solve for \( t \) in the equation:
\( \frac{1}{512} = \frac{1}{2}^{\frac{t}{5.3}} \)
This yields \( t = 47.7 \) years.
Understanding Cobalt-60 decay helps in various areas such as medical oncology with radiation therapy and industrial equipment where it is used for non-destructive testing.