Question

The present level of \({ }^{235} \mathrm{U}\) in naturally occurring uranium ore is \(0.72 \%\). If the half-life of uranium- 235 is \(7.03 \times 10^{8}\) years, how many years ago did naturally occurring uranium contain \(3.00 \%{ }^{235} \mathrm{U}\), the level needed to sustain a chain reaction?

Short answer

1.44 billion years ago.

Step-by-step solution

Identify Key Variables

First, identify the variables and constants essential for the problem: the initial percentage of uranium-235, \(N_0 = 3.00\%\), and the present percentage, \(N = 0.72\%\). Also note the half-life of uranium-235, \(T_{1/2} = 7.03 \times 10^8\) years.

Use the Decay Formula

To find the time elapsed, we use the radioactive decay formula: \[ N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]Substitute known values: \(0.72 = 3.00 \left(\frac{1}{2}\right)^{\frac{t}{7.03 \times 10^8}}\).

Simplify and Solve for Time

Simplify the equation: \[ \frac{0.72}{3.00} = \left(\frac{1}{2}\right)^{\frac{t}{7.03 \times 10^8}} \]Calculate \(\frac{0.72}{3.00} = 0.24\). Then take the logarithm of both sides:\[ \log(0.24) = \frac{t}{7.03 \times 10^8} \log\left(\frac{1}{2}\right) \].

Solve for \(t\)

Rearrange to solve for \(t\):\[ t = \frac{\log(0.24)}{\log(\frac{1}{2})} \times 7.03 \times 10^8 \].Calculate using the logarithmic values: \[ t = \frac{-0.6198}{-0.3010} \times 7.03 \times 10^8 \approx 1.44 \times 10^9 \text{ years} \].

Interpret the Result

The time calculated, \(t \approx 1.44 \times 10^9\) years, indicates that approximately 1.44 billion years ago, the uranium-235 composition was sufficient to support a chain reaction.

Key concepts

Uranium-235

Uranium-235, often abbreviated as U-235, is a crucial isotope of uranium used primarily for its ability to undergo fission. This is a process where the nucleus of an atom splits into smaller parts, releasing a tremendous amount of energy. Unlike Uranium-238, which is more abundant but less reactive, U-235 is capable of sustaining nuclear reactions.
In naturally occurring uranium, U-235 comprises only about 0.72% of the composition. However, to be effective in nuclear reactors or weapons, the percentage must often be increased to about 3% or more through a process known as enrichment. This helped to further our understanding and technological use of nuclear energy.
Understanding the decay properties of U-235 is essential because it determines how long the isotope remains effective for energy production. This is closely tied to its half-life, a foundational concept in nuclear physics.

Half-life

The half-life of a radioactive substance is defined as the time required for half of the atoms in a sample to decay. For Uranium-235, the half-life is approximately 703 million years (\(7.03 \times 10^8 \) years). This incredibly long half-life means that U-235 remains present in the environment for billions of years.
The concept of a half-life is crucial when calculating the age of a sample or understanding the sustainability of a nuclear process. Whether it's determining the age of rocks or the feasibility of a nuclear reactor, knowing the half-life helps us predict how long a material will remain effective before decaying to a less useful form.
For practical uses, the half-life dictates the interval over which radioactive materials must be managed, ensuring safety in transport and handling, and allowing predictions about future reactivity and availability.

Chain Reaction

A chain reaction in nuclear physics occurs when a self-sustaining sequence of reactions happens. In the context of Uranium-235, it's the series of processes where the fission of an atom leads to additional fission events in other nuclei. This is the principle behind both nuclear reactors and atomic bombs.
For a chain reaction to be sustained, a critical mass of fissionable material must be present, often regulated by controlling the amount of U-235 or using moderators to influence neutron speed.
In the exercise, the reference to a 3% concentration of U-235 indicates the threshold level necessary for sustaining such a reaction, allowing for controlled energy release. Without this critical mass or concentration, the reaction stops, underscoring the importance of precise measurements and calculations in nuclear science.

Logarithmic Calculations

Logarithmic calculations are vital when dealing with exponential processes like radioactive decay, where quantities change at rates proportional to their size. In the exercise, logarithms simplify the decay equation to find the time elapsed since the uranium content was at a higher level.
The equation \[ N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \] can be tricky, but by using logarithms, it becomes easier to solve for time \(t\). First, you rearrange the decay equation to \[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]. Then, applying logarithms to both sides allows exponential decay problems to be solved using linear methods:\[ \log\left(\frac{N}{N_0}\right) = \frac{t}{T_{1/2}} \log\left(\frac{1}{2}\right) \].
These steps illustrate the power of logarithms—they convert multiplicative relationships into more manageable linear ones, making complex decay problems like this much simpler to understand and solve.