Problem 36 - Chemistry The Central Science

Question

It takes 4 \(\mathrm{h} 39\) min for a 2.00 - mg sample of radium-230 to decay to 0.25 \(\mathrm{mg.}\) What is the half-life of radium-230?

Step-by-Step Solution

Answer

The half-life of radium-230 is 93 minutes.

Step 1: Convert the given time to minutes

We have 4 hours and 39 minutes. Convert this time to minutes by multiplying the hours by 60 and adding the minutes. Time = (4 hours × 60 minutes/hour) + 39 minutes = 240 + 39 = 279 minutes

Step 2: Write down the decay formula

The decay formula is given by the equation: \[N(t) = N_0 \times 2^{-\frac{t}{T}}\] where - \(N(t)\) is the final mass of the substance at time \(t\), - \(N_0\) is the initial mass of the substance, - \(t\) is the time of decay, - \(T\) is the half-life period.

Step 3: Substitute the given values into the formula

We have \(N(t) = 0.25\,\mathrm{mg}\), \(N_0 = 2.00\,\mathrm{mg}\), and \(t = 279\,\mathrm{min}\). Substitute these values into the decay formula: \[\begin{align*}0.25 &= 2 \times 2^{-\frac{279}{T}}\end{align*}\]

Step 4: Solve for the half-life period \(T\)

Isolate \(2^{-\frac{279}{T}}\) by dividing both sides by 2: \[\begin{align*}2^{-\frac{279}{T}} &= \frac{1}{8}\end{align*}\] Now we can take the logarithm to base 2 on both sides to remove the exponent: \[^{-\frac{279}{T}} = \log_2\left(\frac{1}{8}\right)\] Therefore, \(\,\frac{279}{T}\) will be equal to the result of \(\log_2\left(\frac{1}{8}\right)\): \[\frac{279}{T} = -3\] Now, solve for \(T\): \[T = \frac{279}{3} = 93\]

Step 5: Write the final answer

The half-life of radium-230 is 93 minutes.

Key Concepts

Radioactive Decay

Radioactive decay is a fundamental concept in nuclear chemistry, involving the spontaneous transformation of an unstable atomic nucleus into a more stable one. This process emits radiation in the form of particles or electromagnetic waves and occurs naturally in a variety of elements found on Earth.

Substances that undergo radioactive decay are called radionuclides, and they possess what's known as a half-life, which is the time required for half of the radioactive atoms present in a sample to decay. Radioactive decay is a crucial process in fields such as medicine, where it's used for diagnostic imaging and cancer treatment, as well as in carbon dating, where it helps determine the age of archeological finds.

Decay Formula

Understanding the decay formula is essential in solving problems involving radioactive decay. This formula, represented as
\[N(t) = N_0 \times 2^{-\frac{t}{T}}\],
describes the relationship between the remaining quantity of a substance
(\(N(t)\)) after time
(\(t\)), its initial quantity
(\(N_0\)), and the half-life period
(\(T\)). It’s vital to note that the exponent –\(\frac{t}{T}\) indicates how many half-lives have elapsed. The role of this exponential factor is to calculate how much of the initial sample remains after a given period.

Logarithms in Decay

Logarithms play a crucial role in simplifying the process of solving for the half-life period in decay problems. By applying logarithms, we can transform the exponential decay equation into a linear form, making the half-life easier to calculate.

In our example, taking the logarithm to base 2 of both sides enables us to isolate the half-life
(\(T\)), bypassing the complexity of dealing with exponents directly. This is done because logarithms have the unique property of turning multiplication and powers into addition and multiplication, respectively, thus untangling the variables entwined by the exponential function and allowing for straightforward computation.

Nuclear Chemistry

Nuclear chemistry encompasses the study of radioactive substances and the chemical changes that occur due to nuclear reactions. It involves understanding the processes that alter the composition of atomic nuclei and has vast applications, from generating electricity in nuclear power plants to developing treatments that target cancer cells.

In educational terms, diving into this branch of chemistry equips students with the knowledge of how elements can transform and the effects of such transformations, both scientifically and environmentally. Moreover, it underscores the importance of safety and regulations when dealing with materials that emit radiation, reinforcing the need for responsible practices in this powerful field of science.