Problem 38 - Physics

Question

The isotope \({ }_{79}^{198} \mathrm{Au}\) (atomic mass \(=197.968 \mathrm{u}\) ) of gold has a half-life of \(2.69\) days and is used in cancer therapy. What mass (in grams) of this isotope is required to produce an activity of \(315 \mathrm{Ci}\) ?

Step-by-Step Solution

Answer

The required mass is approximately 1.29 mg of the isotope.

Step 1: Understand the Relationship between Activity, Half-Life, and Mass

Activity (A) is the rate at which a radioactive substance decays. It's related to the number of atoms (N) in a sample and the decay constant (λ) by the equation \( A = \lambda N \). The decay constant \( \lambda \) is linked to the half-life \( t_{1/2} \) by the formula \( \lambda = \frac{0.693}{t_{1/2}} \). Here, we need to find the mass that will give the desired activity.

Step 2: Calculate the Decay Constant

Half-life, \( t_{1/2} \), is given as 2.69 days. First, convert this to seconds since the units for the decay constant \( \lambda \) are per second. There are 86400 seconds in a day, so the half-life in seconds is \( 2.69 \times 86400 = 232416 \) seconds. Calculate \( \lambda \) as follows: \( \lambda = \frac{0.693}{232416} \approx 2.98 \times 10^{-6} \text{ s}^{-1} \).

Step 3: Convert Activity from Ci to Disintegrations per Second (dps)

1 Curie (Ci) is equivalent to \( 3.7 \times 10^{10} \) disintegrations per second. Thus, an activity of 315 Ci corresponds to \( 315 \times 3.7 \times 10^{10} = 1.1655 \times 10^{13} \) dps.

Step 4: Calculate the Number of Atoms Required for the Given Activity

Using \( A = \lambda N \) where \( A = 1.1655 \times 10^{13} \) dps, solve for N: \( N = \frac{A}{\lambda} = \frac{1.1655 \times 10^{13}}{2.98 \times 10^{-6}} \approx 3.91 \times 10^{18} \) atoms.

Step 5: Find the Mass of Gold Isotope

The number of atoms is related to mass by the formula \( N = \frac{m}{A_m} \times N_A \), where \( m \) is the mass, \( A_m \) is the atomic mass (197.968 u), and \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \) atoms/mol). Solve for \( m \): \( m = \frac{N \times A_m}{N_A} = \frac{3.91 \times 10^{18} \times 197.968}{6.022 \times 10^{23}} \approx 1.29 \times 10^{-3} \text{ grams} \).

Key Concepts

Radioactive Decay

Radioactive decay is a fundamental concept in nuclear physics. It refers to the process by which an unstable atomic nucleus loses energy by emitting radiation. During decay, a radioactive element transforms into a different element or a lower energy state. Here, focus is on how often these transformations occur, termed as the decay rate. Understanding decay involves:

• Spontaneous nature: The process is random and doesn't require external energy.
• Types of radiation: Including alpha, beta, and gamma radiation.
• Practices: Often used in health and medicine, particularly in treatments like cancer therapy.

Each decay event is unique, but overall, they collectively follow predictable patterns, allowing scientists to estimate the decay rate.

Half-Life

Half-life is a measure of the time it takes for half of the radioactive atoms in a sample to decay. This crucial concept helps us understand how long a substance remains active. It's important because:

• Consistency: Regardless of how much material you start with, the half-life remains constant.
• Predicting Decay: Allows calculation of how long a radioactive material remains potent.
• Applications: Used to date archaeological findings and manage nuclear waste.

In the problem, gold isotope Half-life formula: \[ t_{1/2} = \frac{0.693}{\lambda} \] where \( \lambda \) is the decay constant.

Decay Constant

The decay constant, often denoted as \( \lambda \), represents the probability per unit time that a nucleus will decay. It's linked intimately with half-life, showing how quickly decay occurs. The smaller the decay constant, the slower the rate of decay. To find the decay constant, use the formula: \[ \lambda = \frac{0.693}{t_{1/2}} \] including the half-life in the equation gives insights into the decay process speed. For example, in the exercise, calculating the decay constant helped determine the decay rate of the \({ }_{79}^{198} \text{Au}\) isotope.

Curie (Ci)

In nuclear physics, Curie is a unit of measurement for radioactivity. It measures how many disintegrations occur per second in a radioactive sample. Named after Marie Curie, one Curie equates to: \[ 1 \text{ Ci} = 3.7 \times 10^{10} \text{ disintegrations per second}\] This unit helps quantify a substance's activity level. The higher the number, the more potent the radioactivity. For instance, a 315 Ci activity implies significant radioactivity, required to understand the gold isotope's activity in the original exercise. Using curie in calculations makes these values manageable, offering a common ground for scientists and health professionals.

Atomic Mass

Atomic mass is the mass of an atom, typically expressed in unified atomic mass units (u) or Daltons. The atomic mass reflects both protons and neutrons in the nucleus—key to determining substance quantity in a sample. For nuclear exercises like the given problem, atomic mass is crucial because:

• Relation to Mass: Calculation of a sample's mass using atomic mass and Avogadro's number.
• Conversion: Helps convert between number of atoms and mass in grams.

In the gold example, the atomic mass 197.968 u aids in translating from a number of gold atoms to their corresponding mass in grams. Understanding atomic mass unites physics, chemistry, and biology in quantifying material amounts.