Question

\({ }_{82}^{214} \mathrm{~Pb}\) has a half-life of \(26.8 \mathrm{~min}\). How many minutes must elapse for \(90.0 \%\) of a given sample of \({ }_{82}^{214} \mathrm{~Pb}\) atoms to decay?

Short answer

Answer: Approximately 88.96 minutes.

Step-by-step solution

Understand the radioactive decay formula

The radioactive decay formula is given by: \(N_t = N_0 \cdot (1/2)^{t/T}\), where \(N_t\) is the amount of a sample remaining after time \(t\), \(N_0\) is the initial amount of the sample, \(T\) is the half-life, and \(t\) is the elapsed time. In this problem, we need to find the time when \(90.0\%\) of the sample has decayed, or when \(10.0\%\) remains.

Set up the equation to solve for time

Since we're given that \(90.0\%\) of the sample has decayed, it means that \(10.0\%\) of the sample remains. Therefore, we have \(N_t = 0.1 N_0\). Our goal is to find the elapsed time (\(t\)), so plug in the values into the radioactive decay formula: \(0.1 N_0 = N_0 \cdot (1/2)^{t/26.8}\)

Solve for the elapsed time (\(t\))

First, divide both sides of the equation by \(N_0\): \(0.1 = (1/2)^{t/26.8}\) Now, we take the logarithm of both sides, so we can isolate \(t\): \(\log(0.1) = \log\left((1/2)^{t/26.8}\right)\) Using the logarithmic property \(\log(a^b) = b \log(a)\), we have: \(\log(0.1) = \frac{t}{26.8} \log\left(\frac{1}{2}\right)\) Next, we isolate \(t\) by multiplying both sides by \(26.8\) and dividing by \(\log\left(\frac{1}{2}\right)\): \(t = \frac{26.8\cdot\log(0.1)}{\log\left(\frac{1}{2}\right)}\)

Calculate the elapsed time

Finally, use a calculator to find the value of \(t\): \(t \approx 88.96 \mathrm{~minutes}\) Therefore, approximately 88.96 minutes must elapse for \(90.0 \%\) of a given sample of \({ }_{82}^{214} \mathrm{~Pb}\) atoms to decay.

Key concepts

Understanding Half-Life

The term 'half-life' is a fundamental concept in the study of radioactive decay, representing the time required for half of a radioactive substance to lose its radioactivity through decay. In simpler terms, after one half-life, there will be half of the original amount of the substance remaining, with the other half having transformed into different atoms.

When calculating decay over multiple half-lives, the remaining amount continues to halve with each interval. This exponential decay is crucial in various fields, including geology for dating rocks and archaeology for carbon dating artifacts. Knowing the half-life of a substance also helps us understand how long it may remain hazardous, influencing containment strategies for radioactive waste management.

Radioactive Decay Formula Explained

In nuclear chemistry, the radioactive decay formula is a mathematical expression that enables us to calculate the remaining quantity of a radioactive substance after a certain time period. The formula is \[\begin{equation} N_t = N_0 \cdot \left(\frac{1}{2}\right)^{t/T} \end{equation}\],where:

• \(N_t\) is the amount of substance left after time \(t\),
• \(N_0\) is the initial amount of the substance,
• \(T\) represents the half-life, and
• \(t\) is the elapsed time in the same unit as the half-life.

Through this formula, one can deduce how much time it will take for a certain amount of the substance to decay, or conversely, how much substance will be left after a given timeframe. This is particularly essential for applications in medical treatments, where precise doses of radioactive substances are necessary for both efficacy and safety.

Logarithmic Properties in Radioactive Decay

Logarithmic properties are indispensable tools when solving equations involving exponential functions, like the radioactive decay formula. A common property used is \[\begin{equation} \log(a^b) = b \cdot \log(a) \end{equation}\],which assists in isolating variables when they appear as an exponent. By applying logarithms to both sides of an equation, one can bring down the exponent as a coefficient, making it possible to solve for the unknown variable.

For example, in the radioactive decay exercise, taking the logarithm of both sides of the equation allowed us to solve for \(t\), the elapsed time. This practical application of logarithms can often be seen not only in nuclear chemistry but also in finance for compound interest calculations and in biology for modeling population growth.