Question

A 0.0100 -g sample of a radioactive isotope with a halflife of \(1.3 \times 10^{9}\) years decays at the rate of \(2.9 \times 10^{4}\) disintegrations per minute. Calculate the molar mass of the isotope.

Short answer

The molar mass of the isotope is approximately 210.53 g/mol.

Step-by-step solution

Understand the formula for decay

The activity (rate of decay) of a radioactive substance is given by the equation \( A = \lambda N \), where \( A \) is the activity, \( \lambda \) is the decay constant, and \( N \) is the number of atoms present.

Determine the decay constant \(\lambda\)

Use the formula for the decay constant \( \lambda = \frac{0.693}{t_{1/2}} \), where \( t_{1/2} = 1.3 \times 10^9 \) years is the half-life. Convert the half-life into minutes for consistency in units. Thus, \( t_{1/2} = 1.3 \times 10^9 \times 365.25 \times 24 \times 60 \approx 6.84 \times 10^{14} \) minutes. Therefore, \( \lambda = \frac{0.693}{6.84 \times 10^{14}} \approx 1.013 \times 10^{-15} \) per minute.

Find the number of atoms \(N\)

Use the decay equation \( A = \lambda N \) to find \( N \), where \( A = 2.9 \times 10^4 \) disintegrations per minute. Rearranging gives \( N = \frac{A}{\lambda} = \frac{2.9 \times 10^4}{1.013 \times 10^{-15}} \approx 2.86 \times 10^{19} \) atoms.

Calculate moles of the isotope

Use Avogadro's number \( 6.022 \times 10^{23} \) to find moles: \( \text{moles} = \frac{2.86 \times 10^{19}}{6.022 \times 10^{23}} \approx 4.75 \times 10^{-5} \) moles.

Calculate the molar mass

Using the sample mass of 0.0100 g, divide by the number of moles to get molar mass: \( M = \frac{0.0100}{4.75 \times 10^{-5}} \approx 210.53 \) g/mol.

Key concepts

Half-life Calculation

The concept of half-life is essential in understanding how quickly a radioactive substance decays. Half-life is the time required for a quantity to reduce to half its initial value. For radioactive substances, this measures how long it takes for half of the atoms in a sample to undergo decay.
In mathematical terms, the decay constant \( \lambda \) is associated with the half-life through the formula:

• \( \lambda = \frac{0.693}{t_{1/2}} \)

Here, \( t_{1/2} \) is the half-life. The constant 0.693 is the natural logarithm of 2, which comes from the exponential decay model. Calculation of the decay constant helps us understand the rate at which isotopes decay, which is crucial for determining the time frame over which significant changes in activity occur.
Multiplying this time duration by the factors that convert years to minutes ensures proper unit consistency, which is key to precise scientific calculations.

Activity of a Radioactive Substance

The activity of a radioactive substance indicates how many atoms disintegrate in a given time period and is expressed as disintegrations per minute (dpm) or per second (Bq).
Activity \( A \) is directly proportional to the number of radioactive atoms \( N \) present and the decay constant \( \lambda \), governed by the equation:

• \( A = \lambda N \)

This equation tells us that the higher the decay constant or number of atoms, the higher the activity.
In practical terms, it gives a sense of how "active" or "energetic" a radioactive sample is, which is vital in applications ranging from medical treatments to nuclear power generation.
Understanding activity allows us to predict how long a substance will remain active and its efficacy in various applications.

Decay Constant

The decay constant \( \lambda \) is a parameter that quantifies the rate of radioactive decay. It is defined as the probability per unit time that an atom will decay and is crucial for understanding the speed at which an atom will decompose.
The decay constant is expressed as:

• \( \lambda = \frac{0.693}{t_{1/2}} \)

The decay constant provides insight into the stability of a radioactive isotope. A higher \( \lambda \) implies a shorter half-life, meaning the isotope decays rapidly, whereas a lower \( \lambda \) suggests a more stable isotope with a longer half-life.
In this exercise, finding \( \lambda \) involved adjusting the half-life to the correct time unit (minutes) for consistency with the given activity, emphasizing the importance of unit conversion in scientific calculations.

Avogadro's Number

Avogadro's number, \( 6.022 \times 10^{23} \), is a fundamental constant that connects the microscopic world of atoms to the macroscopic world of grams and moles. It defines the number of constituent particles (typically atoms or molecules) in one mole of a substance.
When considering moles from a number of atoms \( N \), the relationship is:

• \( \text{moles} = \frac{N}{6.022 \times 10^{23}} \)

Avogadro's number serves as a bridge between atoms and moles, allowing scientists to calculate everything from the mass of a molecule to the amount of a substance in a reaction.
In radioactive decay exercises, like the one described, it plays a crucial role in translating the microscopic decay rate into macroscopic amounts, which can be further utilized to find properties like molar mass of an isotope.