Question

Cobalt-60 is commonly used as a source of \(\beta\) particles. How long does it take for 87.5\(\%\) of a sample of cobalt-60 to decay (the half-life is 5.26 years)?

Short answer

It takes approximately 14.92 years for 87.5% of a cobalt-60 sample to decay.

Step-by-step solution

Determine the Amount of Substance Remaining After Decay

We are given that 87.5% of the cobalt-60 sample has decayed. This means that the remaining undecayed substance is 100% - 87.5% = 12.5%. If we represent this as a decimal, we have 0.125.

Set Up the Half-life Formula

We will use the half-life formula to find the time it takes for 87.5% of the sample to decay. \[ N(t) = N_0 \cdot \left(\dfrac{1}{2}\right)^\frac{t}{T} \] Substitute N(t) with 0.125 * N_0, since 12.5% of the substance remains undecayed. \[ 0.125 N_0 = N_0 \cdot \left(\dfrac{1}{2}\right)^\frac{t}{T} \]

Simplify and Solve for t

Divide both sides of the equation by N_0 to cancel it out. \[ 0.125 = \left(\dfrac{1}{2}\right)^\frac{t}{T} \] Now we need to solve for t. Since the equation involves an exponential function, we can use logarithms to solve for t. The easiest logarithm to use in this case is the natural logarithm (ln) as it allows us to then use the Change of Base formula. \[ \ln{0.125} = \ln{\left(\dfrac{1}{2}\right)^\frac{t}{T}} \] Using the logarithm power rule, we can bring the exponent out in front. \[ \ln{0.125} = \dfrac{t}{T} \cdot \ln{\left(\dfrac{1}{2}\right)} \] Now, divide by the \(\ln{\left(\dfrac{1}{2}\right)} \) to isolate t: \[ t = T \cdot \dfrac{\ln{0.125}}{\ln{\left(\dfrac{1}{2}\right)}} \] Substitute the half-life T with the given value 5.26 years. \[ t = 5.26 \cdot \dfrac{\ln{0.125}}{\ln{\left(\dfrac{1}{2}\right)}} \]

Calculate t

Now, use a calculator to calculate the value of t. \[ t \approx 5.26 \cdot \dfrac{\ln{0.125}}{\ln{\left(\dfrac{1}{2}\right)}} \approx 14.92\, years \] So, it takes approximately 14.92 years for 87.5% of a cobalt-60 sample to decay.

Key concepts

Radioactive decay

Radioactive decay is a natural and spontaneous process by which unstable atomic nuclei lose energy by emitting radiation. This decay occurs in a random manner, leading to the transformation of a parent isotope into a daughter isotope. When discussing radioactive decay, it is essential to understand a few key terms:

• Isotope: Different forms of the same element, with an equal number of protons but a different number of neutrons.
• Half-life: The time it takes for half of a sample of radioactive material to decay. This is a constant specific to each radioactive isotope.
• Beta decay: A type of radioactive decay where a neutron in the nucleus transforms into a proton and emits a beta particle (an electron or positron).

During radioactive decay, no external forces influence the rate at which an isotope decays. The predictability of half-life allows scientists to calculate the rate of decay, which is critical for applications like dating archeological findings or managing medical treatments that involve radioactive substances.

Cobalt-60

Cobalt-60 is a radioactive isotope of cobalt. It is widely used in various sectors due to its radioactive properties. One of its most common applications is as a source of beta particles for radiation therapy in cancer treatment. Cobalt-60 is also used in industrial radiography to inspect metal parts and structures for defects.

• Properties: Cobalt-60 has a half-life of 5.26 years, meaning that every 5.26 years, half of a given sample decays.
• Decay Equation: As cobalt-60 decays, it emits gamma radiation, which is a type of electromagnetic radiation similar to X-rays but with higher energy.
• Safety: Handling cobalt-60 requires stringent safety measures to protect from radiation exposure. It is vital to store and use it in controlled environments with proper shielding.

Due to its predictable half-life and the manner in which it decays, cobalt-60 serves as an excellent example to understand concepts surrounding radioactive materials.

Exponential decay formula

The exponential decay formula describes the process of reducing an amount gradually at a rate proportional to its current value. In the context of radioactive decay, this formula helps predict how much of a substance remains after a certain period. The formula can be expressed as:
\[ N(t) = N_0 \cdot \left(\dfrac{1}{2}\right)^\frac{t}{T} \]

• \(N(t)\): The amount of substance that remains after time \(t\).
• \(N_0\): The original amount of the substance.
• \(t\): The time elapsed, often expressed in years for radioactive substances.
• \(T\): The half-life of the substance.

This formula reflects the nature of half-lives, illustrating how after each half-life period, the substance amount is halved. Using logarithms, students can rearrange this formula to find the time required for a certain percentage of decay, facilitating the understanding and solving of problems involving radioactive decay.