Problem 38 - Thomas Calculus in SI Units

Question

The half-life of polonium is 139 days, but your sample will not be useful to you after \(95 \%\) of the radioactive nuclei present on the day the sample arrives has disintegrated. For about how many days after the sample arrives will you be able to use the polonium?

Step-by-Step Solution

Answer

The polonium will be usable for approximately 600 days.

Step 1: Understand the Concept of Half-Life

Half-life is the time required for an amount to reduce to half its initial value due to decay. Polonium has a half-life of 139 days, meaning after 139 days, only half of the original nuclei remain.

Step 2: Set Up the Exponential Decay Equation

The decay of a radioactive substance is described by the exponential decay formula \( N(t) = N_0 e^{-kt} \), where \( N(t) \) is the amount remaining after time \( t \), \( N_0 \) is the initial amount, and \( k \) is the decay constant.

Step 3: Calculate the Decay Constant

Solve for the decay constant \( k \) using the half-life formula. Since at half-life, \( N(t) = \frac{N_0}{2} \), substitute these into the decay equation: \( \frac{N_0}{2} = N_0 e^{-k imes 139} \). Simplifying, we get \( e^{-139k} = 0.5 \). Solving for \( k \), we find \( k = \frac{\ln(0.5)}{-139} \).

Step 4: Determine When 95% Has Disintegrated

We need to find the time \( t \) when 95% of the nuclei have disintegrated, meaning 5% of the original amount remains: \( N(t) = 0.05N_0 \). Substitute into the decay equation: \( 0.05N_0 = N_0 e^{-kt} \). This simplifies to \( e^{-kt} = 0.05 \), and further to \( -kt = \ln(0.05) \).

Step 5: Solve for Time \( t \)

Using the expression \( -kt = \ln(0.05) \) and the previously calculated \( k \), we find \( t = \frac{\ln(0.05)}{-k} \). Substitute \( k = \frac{\ln(0.5)}{-139} \) into the equation: \( t = \frac{\ln(0.05)}{-\left(\frac{\ln(0.5)}{-139}\right)} \). Calculate \( t \) to find the usable lifespan of the polonium.

Step 6: Calculate and Conclude

Substitute the values into a calculator: \( \ln(0.05) \approx -2.9957 \) and \( \ln(0.5) \approx -0.6931 \). Thus, \( t \approx \frac{2.9957}{0.00499} \approx 600 \) days. Therefore, the polonium will be usable for approximately 600 days after arrival.

Key Concepts

Understanding Half-Life

The concept of half-life is central to understanding radioactive decay. Half-life is the time it takes for half of the radioactive nuclei in a sample to decay. For example, if you have a sample of polonium with a half-life of 139 days, this means every 139 days, half of whatever amount you have left will have decayed. This process continues until the material has decayed to a point where it's no longer effective or useful for its intended purpose.

It's important to note that the half-life is a constant property of the isotope, and doesn't change regardless of the amount of substance you start with.

• If a sample starts with 100 atoms, after one half-life (139 days for polonium), only 50 atoms remain undecayed.
• After two half-lives, just 25 atoms remain.

Understanding how the initial amount decreases over successive half-lives helps determine how long you can use a radioactive sample before it diminishes significantly.

Exploring Radioactive Decay

Radioactive decay is a natural process where an unstable atomic nucleus loses energy by emitting radiation. This process is random for each atom, but predictable for a large number of atoms over time.

To mathematically describe radioactive decay, we use the exponential decay formula: \[N(t) = N_0 e^{-kt}\]where:

• \(N(t)\) is the quantity of a substance that still remains after time \(t\).
• \(N_0\) is the initial quantity of the substance.
• \(e\) is the base of the natural logarithms.
• \(k\) is the decay constant, which is unique to each radioactive substance and determines how quickly it decays.

Since decay is exponential, it means radioactive substances decrease rapidly at first and then level off, making predictions about decay very precise.

The Decay Constant and Its Role

The decay constant \(k\) is a fundamental part of understanding how quickly a radioactive substance decays. It tells us how quickly the nuclei disintegrate over time. The value of \(k\) can be found using the known half-life of the substance.

Using the formula: \[\frac{N_0}{2} = N_0 e^{-k \times \text{half-life}}\]we solve for \(k\) to get: \[k = \frac{\ln(0.5)}{-\text{half-life}}\]Applying this formula, you compute \(k\) for a given substance if its half-life is known, helping predict how the material will behave over time.

• The decay constant is crucial because it allows for calculations of how much of the original substance remains after any given time \(t\).
• A larger decay constant indicates faster decay.
• In our example, with polonium's half-life, \(k\) is used to calculate the time until the substance is no longer useful.

Understanding \(k\) connects the theoretical half-life with practical application, guiding experiments and utilizations of radioactive materials efficiently.