Question

A sample of radioactive substance gave 630 counts per minute and 610 counts per minute at times differing by hour. The decay constant \((\lambda)\) in \(\min ^{-1}\) is given by (a) \(\lambda=\frac{630}{610} \times 60\) (b) \(\mathrm{e}^{60 \mathrm{~A}}=\frac{630}{610}\) (c) \(\lambda=\frac{2.303}{60} \log \frac{610}{630}\) (d) \(\lambda=\frac{2.303}{60} \times \frac{630}{610}\)

Short answer

Option (b) is correct: \( e^{60 \lambda} = \frac{630}{610} \).

Step-by-step solution

Understanding the Options

We are presented with four formula options to determine the decay constant, \( \lambda \), from a radioactive sample measurement. Each formula applies logarithmic or exponential methods to find \( \lambda \) based on the given decay process.

Identifying Time Interval and Decay Formula

The counts decrease from 630 to 610 in one hour, which is crucial in applying the decay formula. The decay process follows the exponential decay law \( N(t) = N_0 e^{-\lambda t} \). Here, \( N_0 = 630 \) and \( N(t) = 610 \) after one hour (or 60 minutes).

Applying Exponential Decay Formula

Using the formula \( N(t) = N_0 e^{-\lambda t} \), we have:\[ 610 = 630 e^{-60 \lambda} \].This can be rearranged to give:\[ e^{60 \lambda} = \frac{630}{610} \].This expression matches option (b) directly from the given choices.

Verifying Option Matching

To ensure no ambiguity, let's cross-check each option:- (a) \( \lambda = \frac{630}{610} \times 60 \) is incorrect because it's not solving for \( \lambda \) through the decay equation.- (b) \( e^{60 \lambda} = \frac{630}{610} \) matches the rearranged formula accurately.- (c) Incorrect, as it employs logarithms with incorrect ratio placement.- (d) Method of multiplication does not reflect the decay law accurately.

Key concepts

Decay Constant

The decay constant is a crucial parameter in radioactive decay, symbolized by \( \lambda \). It describes the rate at which a radioactive substance undergoes decay over time. The decay constant is specific to each substance and is derived from the decay law formula.

To better understand the decay constant, remember the following key points:

• It is measured in inverse time units, like \( \min^{-1} \), indicating the fraction of the radioactive substance that decays per minute.
• A larger decay constant implies a faster rate of decay.
• It helps determine how quickly the count of radioactive particles reduces over time.

The decay constant can be determined through exponential relationships as shown in the problem's option (b) formula. Here, the equation \( e^{60 \lambda} = \frac{630}{610} \) is used to calculate \( \lambda \). This direct use of the exponential decay law helps find \( \lambda \) accurately. Understanding the decay constant helps in predicting the behavior of radioactive substances over specified time intervals.

Exponential Decay

Exponential decay occurs when the quantity of a material decreases at a rate proportional to its current value. In radioactive substances, it refers to the reduction of radioactive particle counts over time. The classic formula of exponential decay is given by \( N(t) = N_0 \cdot e^{-\lambda t} \). Here are some crucial points to keep in mind:

• \( N_0 \) is the initial quantity of the substance.
• \( N(t) \) is the quantity of the substance at time \( t \).
• The exponential part \( e^{-\lambda t} \) determines the rate of decay.

In the exercise, we observed exponential decay through the reduction of counts from 630 to 610 over the span of one hour. The decay law equation is used to set up \( 610 = 630 e^{-60 \lambda} \). Solving this gives an expression for the decay constant, highlighting how exponential decay translates into a specific and measurable rate.

Logarithmic Relationships

Logarithmic relationships are mathematical tools used to deal with exponential equations, especially in the context of decay processes. They allow us to express exponential expressions in terms of simpler algebraic forms. For radioactive decay, taking the natural logarithm of both sides of an exponential equation can be exceptionally useful.

Consider the formula \( e^{60 \lambda} = \frac{630}{610} \). To solve for \( \lambda \), a logarithm can be applied:

• Take the natural log on both sides to eliminate the exponential. This provides \( \ln(e^{60 \lambda}) = \ln(\frac{630}{610}) \).
• Simplify using logarithmic properties: \( 60 \lambda = \ln(\frac{630}{610}) \).

The utility of logarithms lies in their ability to untangle complex exponential functions into linear forms, making it easier to isolate variables like \( \lambda \). This conversion from exponential to logarithmic forms highlights the intricate relationship between these mathematical concepts, especially vital in decay calculations.