Problem 34 - Chemistry: The Central Science

Question

It takes \(5.2\) minutesfor a \(1.000-g\) sample of \({ }^{210} \mathrm{Fr}\) to decay to \(0.250 \mathrm{~g}\). What is the half-life of \({ }^{210} \mathrm{Fr}\) ?

Step-by-Step Solution

Answer

The half-life of francium-210 is \(2.6~\text{minutes}\).

Step 1: Write down the decay formula

Write down the decay formula: \( N(t) = N_0 \times (1/2)^{\frac{t}{T}}\). Remember that: - \(N(t)\) is the remaining mass after decay, - \(N_0\) is the initial mass, - \(t\) is the time taken for decay, and - \(T\) is the half-life of the element.

Step 2: Insert the known values

We are given: - Initial mass \(N_0 = 1.000~\text{g}\), - Remaining mass after decay \(N(t) = 0.250~\text{g}\), - Time taken for the decay \(t = 5.2~\text{minutes}\). Insert these values into the decay formula: \(0.250 = 1.000 \times (1/2)^{\frac{5.2}{T}}\).

Step 3: Solve the equation for T

We are now left with an equation with one unknown, \(T\), the half-life of the element: \(0.250 = 1.000 \times (1/2)^{ \frac{5.2}{T}}\) First, divide both sides by 1.000 (which doesn't change the equation): \(0.250 = (1/2)^{\frac{5.2}{T}}\) Take the natural logarithm (ln) of both sides of the equation to get rid of the exponent: \(ln(0.250) = ln \left[ (1/2)^{\frac{5.2}{T}} \right]\) Now, use the property of logarithms that states \(ln(a^b) = b \times ln(a)\): \(ln(0.250) = \frac{5.2}{T} \times ln(1/2)\) Now, solve for \(T\): \(T = \frac{5.2 \times ln(1/2)}{ln(0.250)}\)

Step 4: Calculate the half-life

Now, plug the numbers into the equation and calculate the half-life \(T\): \(T = \frac{5.2 \times ln(1/2)}{ln(0.250)} = 2.6~\text{minutes}\) Thus, the half-life of francium-210 is 2.6 minutes.

Key Concepts

Radioactive Decay

Understanding radioactive decay is crucial when studying nuclear chemistry and physics. Radioactive decay is a random process where an unstable atomic nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves. This process transforms the original element into another element or a different isotope of the original element. The term 'half-life' is used to describe the time required for half of the radioactive atoms in a sample to decay. The half-life is consistent for a given isotope, which means it does not depend on the amount of the substance or its chemical state but is a unique characteristic of the isotope itself.

To visualize it, imagine a large group of identical, unstable atoms. After one half-life passes, half of these atoms will have decayed into a more stable form. After another half-life, half of the remaining unstable atoms will decay, leaving a quarter of the original amount, and so forth.

Decay Formula

The decay formula is a mathematical representation of radioactive decay. The formula is written as
\[N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T}}\]
where \(N(t)\) represents the amount of substance remaining after time \(t\), \(N_0\) is the initial amount of the substance, \(T\) is the half-life of the substance, and \(t\) is the elapsed time.

This formula is based on the principle of exponential decay, which implies that the quantity of a radioactive element decreases exponentially over time. The fraction \(\frac{1}{2}\) is raised to the power of the ratio of time elapsed divided by the half-life, representing the successive halving of the substance amount at each half-life interval. By plugging in the known values into this formula, we can solve for any missing variable, which in this exercise, is the half-life of the element.

Logarithms in Chemistry

Logarithms are an indispensable tool in chemistry, especially when dealing with exponential processes such as radioactive decay. A logarithm is the power to which a number, called the base, must be raised to produce another number. In the context of the decay formula, logarithms help us to solve the equation when the unknown variable is in the exponent.

For example, the natural logarithm function (denoted as \(ln\)) is often used. It has a base of \(e\), which is an irrational constant approximately equal to 2.71828. The logarithmic property \(ln(a^b) = b \times ln(a)\) allows us to move the exponent in front of the logarithm, making it possible to isolate and solve for the exponent, as illustrated in our exercise. This maneuvering is a crucial step in calculating the half-life when dealing with an equation that has an exponential form.

Exponential Decay

Exponential decay is a pattern of decrease in which a quantity diminishes at a rate proportional to its current value. This behavior is described by an exponential function, which consistently halves the remaining amount of a substance over equal intervals of time—the half-lives. In the context of radioactive decay, this means that the rate of decay is fastest at the start and decreases over time as the amount of the substance decreases.

Mathematically, exponential decay is represented by a decaying exponential function, where the base of the exponent is less than one (such as \(\frac{1}{2}\) in the decay formula). The half-life calculation involves understanding how exponential decay operates because it provides the framework for quantifying exactly how quickly a substance is diminishing over time. Understanding this principle allows scientists to determine the age of substances, predict the stability of elements, and safely manage radioactive materials.