Rate of Radioactive Decay: Videos & Practice Problems

In the study of chemical kinetics, particularly regarding radioactive processes, it is essential to understand that these reactions adhere to a first-order rate law. The integrated rate law for radioactive decay can be expressed using the following equation:

$$ \ln(n) = -kt + \ln(n_0) $$

In this equation, \( n \) represents the final concentration of the radioactive nuclei, while \( n_0 \) (or \( n_{\text{naught}} \)) denotes the initial concentration. The variable \( k \) is referred to as the decay constant, which retains the same time-inverse units as a rate constant, meaning that if \( k \) is expressed in days^-1, time must also be measured in days. It is crucial to ensure that the units of time and the decay constant are consistent.

It is important to note that the term "concentration" in this context is not limited to molarity; it can also refer to disintegrations per second or other relevant measures in radioactive processes.

This integrated rate law can be related to the equation of a straight line, \( y = mx + b \), where:

• \( y \) corresponds to \( \ln(n) \) (the natural logarithm of the final concentration),
• \( m \) is the slope, which equals \( -k \),
• \( x \) represents time, and
• \( b \) is \( \ln(n_0) \) (the natural logarithm of the initial concentration).

Since the slope is negative, it indicates a decrease in concentration over time. When plotting \( \ln(n) \) against time, the graph will show a descending line, reflecting the negative slope of \( -k \). The slope can also be defined as the change in \( y \) (natural log of concentration) over the change in \( x \) (time), or:

$$ \text{slope} = \frac{\Delta y}{\Delta x} = \frac{\Delta \ln(n)}{\Delta t} $$

Understanding and applying this radioactive integrated rate law is fundamental, as it will frequently be utilized in various calculations and analyses related to radioactive decay. Familiarity with the variables and their meanings is crucial for effective application in practical scenarios.

To determine the time it takes for the concentration of astatine-210 to decrease from \(9.3 \times 10^5\) disintegrations per second to \(2.7 \times 10^4\) disintegrations per second, we can use the integrated rate law for radioactive decay. The equation is expressed as:

\[\ln\left(\frac{[C_f]}{[C_i]}\right) = -kt\]

In this context, \([C_f]\) represents the final concentration, \([C_i]\) the initial concentration, and \(k\) is the decay constant. Here, we have:

\([C_f] = 2.7 \times 10^4\) disintegrations per second

\([C_i] = 9.3 \times 10^5\) disintegrations per second

\(k = 0.086\) hours^-1

Substituting these values into the equation gives:

\[\ln(2.7 \times 10^4) - \ln(9.3 \times 10^5) = -0.086t\]

Calculating the natural logarithms:

\[\ln(2.7 \times 10^4) \approx -3.589\]

\[\ln(9.3 \times 10^5) \approx 14.007\]

Thus, the equation simplifies to:

\[-3.589 - 14.007 = -0.086t\]

Combining the logarithmic values results in:

\[-17.596 = -0.086t\]

To isolate \(t\), divide both sides by \(-0.086\):

\[t = \frac{17.596}{0.086} \approx 204.6 \text{ hours}\]

Since the question requires the answer in minutes, we convert hours to minutes by multiplying by 60:

\[t \approx 204.6 \times 60 \approx 12276 \text{ minutes}\]

Therefore, the time it takes for the concentration of astatine-210 to decrease from \(9.3 \times 10^5\) disintegrations per second to \(2.7 \times 10^4\) disintegrations per second is approximately 12276 minutes.