Problem 26 - University Physics with Modern Physics

Question

As a health physicist, you are being consulted about a spill in a radiochemistry lab. The isotope spilled was 500\(\mu C\) of \(^{131} \mathrm{Ba}\), which has a half-life of 12 days. (a) What mass of \(^{131} \mathrm{Ba}\) was spilled? (b) Your recommendation is to clear the lab until the radiation level has fallen 1.00\(\mu\) Ci. How long will the lab have to be closed?

Step-by-Step Solution

Answer

A mass of 6.02 ng of Ba-131 was spilled. The lab needs to be closed for 137.3 days.

Step 1: Identify Known Values

We are given that the activity of the spill is 500 \( \mu \text{Ci} \), and the half-life of \( ^{131} \text{Ba} \) is 12 days.

Step 2: Determine the Relationship Between Activity and Mass

The activity (\( A \)) of a radioactive substance is related to the number of atoms (\( N \)) and the decay constant (\( \lambda \)) by the formula: \[ A = \lambda N \]The decay constant is related to the half-life (\( T_{1/2} \)) by:\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]

Step 3: Calculate Number of Atoms

First, we convert 500 \( \mu \text{Ci} \) to International System units:\[ 500 \mu\text{Ci} = 500 \times 3.7 \times 10^{4} \text{ disintegrations per second} = 1.85 \times 10^{7} \text{ Bq} \]Now calculate \( N \):Using \[ \lambda = \frac{\ln(2)}{12 \times 86400} \, \text{s}^{-1} = 6.68 \times 10^{-7} \, \text{s}^{-1} \] \[ N = \frac{A}{\lambda} = \frac{1.85 \times 10^{7}}{6.68 \times 10^{-7}} \approx 2.77 \times 10^{13} \text{ atoms} \]

Step 4: Calculate Mass of Spilled Ba-131

The number of atoms \( N \) is related to the mass (\( m \)) by the formula:\[ N = \frac{m}{m_a} \times N_A \]where \( m_a \) is the molar mass of \( ^{131} \text{Ba} \) (131 g/mol) and \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \text{ mol}^{-1} \)). Solving for \( m \):\[ m = \frac{N \times m_a}{N_A} = \frac{2.77 \times 10^{13} \times 131}{6.022 \times 10^{23}} \approx 6.02 \times 10^{-9} \text{ g} \]

Step 5: Calculate Time for Radiation Level to Fall (Activity Decay)

The decay of activity over time can be expressed by the equation:\[ A(t) = A_0 \times e^{-\lambda t} \]We want to find \( t \) when \( A(t) = 1 \mu \text{Ci} = 1 \times 3.7 \times 10^{4} \text{ Bq} \).Starting with 1.85 \times 10^{7} Bq:\[ 1.85 \times 10^{7} \times e^{-6.68 \times 10^{-7} t} = 3.7 \times 10^{4} \]Divide both sides by 1.85 \times 10^{7}:\[ e^{-6.68 \times 10^{-7} t} = \frac{3.7 \times 10^{4}}{1.85 \times 10^{7}} \]\[ e^{-6.68 \times 10^{-7} t} \approx 2 \times 10^{-3} \]Taking logarithms:\[ -6.68 \times 10^{-7} t = \ln(2 \times 10^{-3}) \]\[ t \approx \frac{\ln(2 \times 10^{-3})}{-6.68 \times 10^{-7}} \approx 137.3 \text{ days} \]

Key Concepts

Half-Life

Half-life is a crucial concept in understanding radioactive decay. It is defined as the time required for half of the radioactive nuclei in a sample to decay. This concept helps determine how quickly a radioactive material will reduce its activity over time. In the context of the exercise, we have a half-life of 12 days for the isotope \(^ {131} \text{Ba}\).
When dealing with real-world radioactive materials, knowing the half-life allows scientists and health physicists to calculate the safety and stability of their environment over time. After every 12 days, only half of the remaining radioactive nuclei would still be active. If you start with 100 units of a substance, after one half-life, you would have 50 units remaining, then 25 after the next half-life, and so forth.
Understanding half-life is essential for planning safe handling, storage, and disposal of radioactive substances.

Decay Constant

The decay constant, symbolized by \( \lambda \), is another fundamental concept in radioactive decay. It represents the probability per unit time that an atom of a radioactive isotope will decay. The relationship between the decay constant and half-life is given by the formula:\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]where \( T_{1/2} \) is the half-life.
In this exercise, the decay constant helps in calculating the initial and future activity of \(^ {131} \text{Ba}\). A smaller decay constant indicates a slower decay process, while a larger decay constant means the material decays quickly. Given that the half-life of \(^ {131} \text{Ba}\) is 12 days, the decay constant would be 6.68 \times 10^{-7} \text{ s}^{-1}, which is used to predict how the activity level changes over time and estimate the time until it decreases to a safe level.
Knowing the decay constant is invaluable for making informed decisions related to radiological safety.

Becquerel (Bq)

Becquerel (Bq) is the SI unit of radioactivity. It measures the amount of radioactive decays per second. One Becquerel corresponds to one decay per second. This unit is named after Henri Becquerel, who discovered radioactivity.
In this exercise, the activity of the spill was first given in microcuries (\( \mu \text{Ci} \)), needing conversion to Bq to use within standard calculations. The conversion factor is 1 \( \mu \text{Ci} = 3.7 \times 10^4 \text{ Bq} \). By converting the activity into Bq, we got 1.85 \times 10^7 Bq, which was essential for further calculations.
Using Bq as a consistent unit of measure simplifies the computation across different parameters like half-life and decay constant. It also allows for internationally standard discussions and experiments on radioactive decay.

Molar Mass

Molar mass is a concept from chemistry that indicates the mass of one mole of a substance, usually measured in grams per mole \( \text{g/mol} \). It connects the micro level of individual atoms to the macro level where the weights of substances are considered.
In the exercise, the molar mass of \(^ {131} \text{Ba}\) is given as 131 g/mol. This value is used to connect the number of radioactive atoms, calculated from the activity and decay constant, to the mass of the material spilled.
The relationship utilized is:\[ N = \frac{m}{m_a} \times N_A \]where \( N \) is the number of atoms, \( m \) is the mass, \( m_a \) is the molar mass, and \( N_A \) is Avogadro's number (6.022 \times 10^{23} \text{ mol}^{-1}).

• This connection through molar mass allows for translating from microscopic atomic scales to practical scales of grams, making the study of materials more tangible and applicable in fields such as health physics.