Problem 41 - Calculus

Question

In the early 1960 s, radioactive strontium-90 was released during atmospheric testing of nuclear weapons and got into the bones of people alive at the time. If the half-life of strontium-90 is 29 years, what fraction of the strontium-90 absorbed in 1960 remained in people's bones in \(2010 ?\) [Hint: Write the function in the form \(\left.Q=Q_{0}(1 / 2)^{t / 29} .\right]\)

Step-by-Step Solution

Answer

About 31.2% of the original strontium-90 remained in 2010.

Step 1: Understand the Formula

The formula given is \( Q = Q_0 \left(\frac{1}{2}\right)^{t/29} \). This is used to calculate the remaining quantity \( Q \) of a substance after time \( t \), given its initial quantity \( Q_0 \) and half-life (in this case, 29 years).

Step 2: Identify Known Values

In this exercise, \( Q_0 \) represents the initial amount of strontium-90 (which can be any arbitrary starting amount, since we are finding a fraction), the half-life is 29 years, and \( t \) is the number of years from 1960 to 2010. Calculate \( t = 2010 - 1960 = 50 \) years.

Step 3: Substitute Values into the Formula

Substitute \( t = 50 \) into the formula to find the fraction remaining:\[ Q = Q_0 \left(\frac{1}{2}\right)^{50/29} \]

Step 4: Simplify the Exponent

Calculate the exponent \( 50/29 \):\[ \frac{50}{29} \approx 1.724 \]

Step 5: Calculate the Remaining Fraction

Substitute the exponent back into the decay formula:\[ Q = Q_0 \left(\frac{1}{2}\right)^{1.724} \]Calculate \( \left(\frac{1}{2}\right)^{1.724} \approx 0.312 \). Thus, approximately 31.2% of the original strontium-90 remained.

Key Concepts

Half-Life

In the realm of radioactive decay, the term "half-life" is crucial. It refers to the time it takes for half of a radioactive substance to decay. In our case, strontium-90 has a half-life of 29 years. This means every 29 years, half of the original strontium-90 isotopes will transform into another element through radioactive decay.

This concept is crucial because it helps predict how long a radioactive substance remains active or hazardous. With each half-life, the quantity of the radioactive substance is halved, making it less significant over time. For our strontium-90 scenario, understanding half-life permits us to calculate how much of the substance remains over a period – in this exercise, over 50 years from 1960 to 2010.

Given a known half-life, the process of calculating the remaining quantity after a period is a straightforward application of exponential decay, using the equation:

• \( Q = Q_0 \left(\frac{1}{2}\right)^{t/29} \)

Exponential Decay

Exponential decay describes situations where quantities decrease at rates proportional to their current value. It's a common mathematical model for processes like radioactive decay, where substances decrease steadily over time. Unlike linear decay, which implies a constant rate of decrease, exponential decay is all about proportions.

In radioactive decay, the concept is captured within the calculation of remaining quantities over time, as shown by the formula we used:

• \( Q = Q_0 \left( \frac{1}{2} \right)^{t/29} \)

Here, \( Q_0 \) is your starting amount, and \( Q \) is what remains after time \( t \). In this formula, \( \left( \frac{1}{2} \right)^{t/29} \) signifies the "fraction remaining" after a given period, where \( t \) is measured in years.

The power of the mathematical base \( \frac{1}{2} \) indicates how each substance holds its halfway point over each half-life. In our example, substituting \( t = 50 \) yields approximately 31.2% of the original strontium-90 still present by 2010, illustrating the effectiveness of an exponential decay model in predicting long-term reduction.

Strontium-90

Strontium-90 is a radioactive isotope produced by nuclear fission of uranium and plutonium in nuclear reactors and during nuclear explosions. It's one of the notable fission products due to its half-life of 29 years. This relatively long half-life allows it to persist in the environment and accumulate over time, leading to potential health risks.

When strontium-90 is absorbed into the human body, it behaves similarly to calcium, often lodging in bones and disrupting normal bone growth and repair. This biological mimicry is why radioactive isotopes like strontium-90 are particularly concerning, as they can remain and cause damage for a prolonged period.

The atmospheric tests in the early 1960s released significant quantities of strontium-90, which dispersed widely. Understanding how much remains over decades can be critical for assessing long-term exposure risks and guiding safety regulations. By applying principles of half-life and exponential decay, we pinpoint that about 31.2% still resided in the bones of people after 50 years from the initial exposure.