Problem 30 - Chemistry and Chemical Reactivity

Question

Copper acetate containing \(^{64} \mathrm{Cu}\) is used to study brain tumors. This isotope has a half-life of 12.7 h. If you begin with \(25.0 \mu \mathrm{g}\) of \(^{64} \mathrm{Cu},\) what mass in micrograms remains after \(64 \mathrm{h} ?\)

Step-by-Step Solution

Answer

After 64 hours, approximately 0.775 micrograms of \(^{64} \mathrm{Cu}\) remain.

Step 1: Understand the Half-Life Formula

The amount of radioactive material remaining after time \( t \) is given by the formula: \\[N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}\]where \( N_0 \) is the initial amount, \( T_{1/2} \) is the half-life, and \( t \) is the time elapsed.

Step 2: Identify Given Values

We know that the initial mass \( N_0 = 25.0 \mu \mathrm{g} \), half-life \( T_{1/2} = 12.7 \) hours, and elapsed time \( t = 64 \) hours. We need to find \( N(t) \).

Step 3: Calculate the Number of Half-Lives

Determine how many half-lives have passed by dividing the elapsed time by the half-life: \\[ \text{Number of half-lives} = \frac{64 \text{ hours}}{12.7 \text{ hours/half-life}} \approx 5.04 \]

Step 4: Calculate Remaining Mass

Substitute the number of half-lives into the formula to calculate the remaining mass: \\[N(t) = 25.0 \mu \mathrm{g} \times \left(\frac{1}{2}\right)^{5.04} \approx 25.0 \mu \mathrm{g} \times 0.031 \approx 0.775 \mu \mathrm{g} \]

Step 5: Conclude the Solution

The mass of \(^{64}\mathrm{Cu}\) remaining after 64 hours is approximately \(0.775 \mu \mathrm{g}\).

Key Concepts

Radioactive Decay

Radioactive decay is a natural process in which an unstable atomic nucleus loses energy by emitting radiation. Over time, this process results in the transformation of an unstable element into a more stable one. This phenomenon is characterized by the half-life, which is the time required for half of the radioactive atoms in a sample to decay.

• The decay process is random, meaning it's impossible to predict which particular atom will decay at a given moment.
• The rate of decay is constant over time for a given isotope.

The half-life formula, \[N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}\], helps calculate the remaining quantity of a radioactive substance. Here, \(N_0\) is the initial amount, \(t\) is the time elapsed, and \(T_{1/2}\) is the half-life.
In the case of \(^{64} \mathrm{Cu}\), the half-life is 12.7 hours, meaning every 12.7 hours, half of the \(^{64} \mathrm{Cu}\)atoms decay.

Copper Acetate

Copper acetate is a compound composed of copper and acetic acid. It appears as blue or green crystals and is used in various applications, including studies involving radioactive isotopes.

• Copper acetate can facilitate the delivery of radioactive copper isotopes, such as \(^{64} \mathrm{Cu}\), into biological systems.
• This property makes it useful in medical imaging and cancer research, particularly for studying brain tumors.

In research, \(^{64} \mathrm{Cu}\) attached to copper acetate is often used as a tracer for detecting metabolic activity or changes within the body. This helps in diagnosing conditions such as cancer by highlighting areas with increased activity, often indicative of tumor growth.
Understanding the process of decay is crucial in determining the appropriate dosage and timing when using such compounds in medical settings.

Isotopes

Isotopes are variants of a chemical element that have the same number of protons but differ in the number of neutrons. This difference in neutron number gives isotopes distinct physical properties, although they retain similar chemical behavior.

• Stable isotopes do not change over time, whereas radioactive isotopes undergo decay.
• Isotopes are widely used in research, medicine, and industry.

A well-known isotope is \(^{64} \mathrm{Cu}\), a form of copper. It is radioactive and has a half-life of 12.7 hours. Used in medical imaging, \(^{64} \mathrm{Cu}\) helps trace biological processes within the body due to its decay properties.
Different isotopes of the same element can vary significantly in stability and application. This makes them invaluable tools for scientists and healthcare professionals in understanding and diagnosing complex biological and physical phenomena.