Problem 48 - Chemistry and Chemical Reactivity

Question

Scandium occurs in nature as a single isotope, scandium-45. Neutron irradiation produces scandium- \(46,\) a \(\beta\) emitter with a half-life of 83.8 days. If the initial activity is \(7.0 \times 10^{4} \mathrm{dpm},\) draw a graph showing disintegrations per minute as a function of time during a period of 1 year.

Step-by-Step Solution

Answer

Graph activity vs time using exponential decay; \( \lambda = 0.00827 \text{ day}^{-1} \), \( t = 365 \text{ days} \).

Step 1: Understand the Exponential Decay Formula

The radioactive decay process is governed by an exponential decay formula given by \( A(t) = A_0 e^{-\lambda t} \), where \( A(t) \) is the activity at time \( t \), \( A_0 \) is the initial activity, and \( \lambda \) is the decay constant.

Step 2: Calculate the Decay Constant

The decay constant \( \lambda \) can be calculated using the formula \( \lambda = \frac{\ln(2)}{t_{1/2}} \), where \( t_{1/2} \) is the half-life. For scandium-46, \( t_{1/2} = 83.8 \) days, so \( \lambda = \frac{0.693}{83.8} \approx 0.00827 \text{ day}^{-1} \).

Step 3: Convert Time to Consistent Units

The problem requires the activity as a function of time over one year. Convert the year into days to maintain consistent time units. One year has approximately 365 days.

Step 4: Calculate Activity Over Time Intervals

Using the formula \( A(t) = A_0 e^{-\lambda t} \), compute the activity at various time intervals within the one-year period. For example, calculate \( A(0), A(30), A(60), \ldots, A(365) \). Use \( A_0 = 7.0 \times 10^4 \) dpm.

Step 5: Create the Graph

Plot these calculated activities against time on a graph with time (in days) on the x-axis and disintegrations per minute (dpm) on the y-axis. The curve will exhibit an exponential decay pattern.

Key Concepts

Scandium Isotopes

Scandium isotopes are variations of the element scandium that differ in the number of neutrons in their nuclei. Scandium naturally occurs in the form of one stable isotope, scandium-45. This means it has 21 protons and 24 neutrons.
When scandium-45 is subjected to neutron irradiation, it transforms into scandium-46. This is a radioactive isotope and acts very differently due to its unstable nature.
Scandium-46 is known for being a beta (\(\beta\)) emitter, indicating that it releases beta particles during its decay process. This property is often used in scientific experiments and applications, where its predictable behavior is crucial for measurements and reactions.

Exponential Decay Formula

The exponential decay formula is central to understanding radioactive decay. It's written as \[ A(t) = A_0 e^{-\lambda t} \] where:

• \( A(t) \) is the remaining activity at time \( t \).
• \( A_0 \) is the initial activity, which is where decay begins.
• \( e \) is the base of the natural logarithm, approximately equal to 2.718.
• \( \lambda \) is the decay constant, a crucial figure derived from the half-life.

The formula represents how a quantity decreases at a rate proportional to its current value, typical in radioactive decay. Being an exponential function, it visually manifests as a steady decline on a graph, reflecting a drop where larger amounts decay faster than smaller ones.

Half-life Calculation

Half-life is a vital concept in radioactivity, explaining how long it takes for half of a given amount of a radioactive substance to decay. For scandium-46, the half-life is 83.8 days. It’s calculated using the formula: \[ \lambda = \frac{\ln(2)}{t_{1/2}} \] where:

• \( t_{1/2} \) is the half-life.
• \( \ln(2) \approx 0.693 \) is the natural logarithm of 2.
• \( \lambda \) is the decay constant.

By replacing \( t_{1/2} \) with 83.8 days, we find \( \lambda \approx 0.00827 \text{ day}^{-1} \). Half-life helps in understanding how quickly a radioactive isotope like scandium-46 loses activity over time. It's essential for predicting the behavior of radioactive materials.

Activity over Time

Tracking activity over time is paramount in radioactive studies. For scandium-46 with an initial activity of \( 7.0 \times 10^4 \) dpm, the exponential decay formula \( A(t) = A_0 e^{-\lambda t} \) is used to study how activity decreases. By inserting values such as \( A_0 = 7.0 \times 10^4 \) and \( \lambda = 0.00827 \text{ day}^{-1} \), activity can be computed for various time intervals, like 0 days, 30 days, and so on up to 365 days.
This provides a sequence of data points representing the decline in activity over time. When plotted on a graph, these points form a characteristic exponential decay curve. This method lets us visualize how quickly the activity diminishes, aiding researchers in determining when the isotope will be too inactive for further use.