Question

The activity of a radioactive sample is measured as \(\mathrm{N}_{0}\) counts per minute at \(\mathrm{t}=0\) and \(\left(\mathrm{N}_{0} / \mathrm{e}\right)\) counts Per minute at \(\mathrm{t}=5 \mathrm{~min} .\) The time (in min) at which activity reduces to half its value is (A) log e \((2 / 5)\) (B) \(5 \log _{10} 2\) (C) \(5 \log _{\mathrm{e}} 2\) (D) \(\log _{10}^{(2 / 5)}\)

Short answer

The time at which the activity reduces to half of its initial value is \(5\ln2\) minutes. The correct answer is (C) \(5 \log_{e} 2\).

Step-by-step solution

Understand the decay law formula

The decay law formula is given by \[N(t) = N_{0} \cdot e^{-\lambda t}\], where - \(N(t)\) is the remaining activity at time \(t\), - \(N_{0}\) is the initial activity, - \(\lambda\) is the decay constant, and - \(t\) is the elapsed time. We are given \(N_0\) and \(N(t)\) at different points in time. We will use this information to determine the decay constant \(\lambda\).

Calculate the decay constant \(\lambda\)

We have the following information: - At \(t = 0\), the sample activity is \(N_0\). - At \(t = 5\) minutes, the sample activity is \(N_0 / e\). Using the decay law formula for \(t = 5\) minutes, we get: \[\frac{N_0}{e} = N_0 \cdot e^{-\lambda (5)}\] Divide both sides by \(N_0\): \[\frac{1}{e} = e^{-5\lambda}\] Now take the natural logarithm of both sides: \[-1 = -5\lambda\] Solving for \(\lambda\), we get: \[\lambda = \frac{1}{5}\]

Calculate the half-life time

The half-life time, denoted as \(t_{1/2}\), is the time required for the activity to reduce to half its initial value. Mathematically, this means that \(N(t_{1/2}) = \frac{N_0}{2}\). Using the decay law formula, we can write: \[\frac{N_0}{2} = N_0 \cdot e^{-\lambda t_{1/2}}\] Divide both sides by \(N_0\): \[\frac{1}{2} = e^{-\frac{1}{5}t_{1/2}}\] Take the natural logarithm of both sides: \[-\ln2 = -\frac{1}{5}t_{1/2}\] Solving for \(t_{1/2}\), we get: \[t_{1/2} = 5\ln2\] So, the time at which the activity reduces to half of its initial value is \(5\ln2\) minutes. The correct answer is (C) \(5 \log_{e} 2\).

Key concepts

Decay Constant

The decay constant, represented by \(\lambda\), is a crucial parameter in understanding how a radioactive substance decays over time. It gives us information about the rate at which the substance loses its activity. The higher the decay constant, the faster the decay rate of the substance.
In the context of the exercise, the decay constant can be calculated by utilizing the exponential decay formula. Given that the activity of the sample decreases from \(N_0\) to \(\frac{N_0}{e}\) in 5 minutes, we can determine that \(\lambda = \frac{1}{5}\). This indicates that for this particular radioactive sample, the rate of decay results in the activity being reduced by a factor of \(e\) every 5 minutes. Understanding the decay constant is essential for predicting future decays and understanding the stability of the radioactive material.

Half-Life

Half-life, often denoted as \(t_{1/2}\), is the time required for a radioactive substance to reduce to half of its initial activity. It's a practical way to grasp how quickly a material becomes less radioactive. In the context of our problem, knowing the half-life allows us to predict when the activity of the sample will reach a certain threshold.
Using the exponential decay formula, we can establish the relationship: \[\frac{N_0}{2} = N_0 \cdot e^{-\lambda t_{1/2}}\]Simplifying by dividing each side by \(N_0\), we get \[\frac{1}{2} = e^{-\frac{1}{5}t_{1/2}}\]By taking the natural logarithm on both sides, solving gives us the half-life as \[t_{1/2} = 5\ln2\] minutes. This result shows that every 5\(\ln2\) minutes, the radioactivity of the sample will decrease by half, highlighting how knowledge of half-life helps in planning and safety measurements around radioactive material.

Exponential Decay Formula

The exponential decay formula is central in modeling how radioactive decay occurs over time and is represented as \[N(t) = N_0 \cdot e^{-\lambda t}\]This equation describes how the quantity of radioactive substance changes as time progresses. Here \(N(t)\) is the remaining activity after time \(t\), while \(N_0\) is the initial activity.
Let's breakdown how this formula applies to the problem at hand:

• The activity at \(t=0\) is given as \(N_0\).
• At \(t=5\) minutes, the activity is \(\frac{N_0}{e}\), indicating decay.

Substituting these conditions into the exponential decay formula allows us to solve for the decay constant and understand the rate of decay." The exponential nature captured by the formula emphasizes how the decay process isn't linear but speeds up or slows down dramatically over time.

Natural Logarithm

The natural logarithm, often written as \(\ln\), is an important concept in the study of exponential growth and decay, commonly used in radioactive decay calculations. It represents the logarithm to the base \(e\) (approximately 2.718). This mathematical tool helps unravel the equations involving exponentials, especially when we seek solutions for time or decay constants.
In the exercise, we use the natural logarithm to solve for both the decay constant \(\lambda\) and the half-life \(t_{1/2}\). By involve \(\ln\) in these calculations, it's possible to transform an exponential equation into a linear form, which simplifies problem solving. For example, taking the natural log of both sides of \(e^{-\lambda t} = \frac{1}{e}\) makes it easier to isolate and calculate \(\lambda\). Such transformations are pivotal in comprehending how quickly substances decay and making predictions based on decay rates.