Problem 127 - A Complete Resource Book in Chemistry for JEE Main

Question

How much time is required for a \(5.75\)-mg sample of \({ }^{51} \mathrm{Cr}\) to decay to \(1.50 \mathrm{mg}\) if it has a half-life of \(27.8\) days? (a) \(5.39\) days (b) \(2.69\) days (c) \(53.9\) days (d) \(5.49\) days

Step-by-Step Solution

Answer

The time required for the sample to decay to 1.50 mg is approximately 53.9 days.

Step 1: Understand Decay Formula

The decay of a radioactive substance is modeled by the exponential decay formula: \( N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \). Here, \( N(t) \) is the remaining quantity, \( N_0 \) is the initial quantity, \( T_{1/2} \) is the half-life of the substance, and \( t \) is the time elapsed.

Step 2: Set Up the Equation

We need to find the time \( t \) for the mass of \( ^{51}\mathrm{Cr} \) to decay from \( N_0 = 5.75 \) mg to \( N(t) = 1.5 \) mg, with a half-life \( T_{1/2} = 27.8 \) days. The equation becomes \( 1.5 = 5.75 \left(\frac{1}{2}\right)^{\frac{t}{27.8}} \).

Step 3: Solve for \( \frac{t}{27.8} \)

Divide both sides by \( 5.75 \) to isolate the exponential term: \( \frac{1.5}{5.75} = \left(\frac{1}{2}\right)^{\frac{t}{27.8}} \). Simplifying gives \( 0.2609 \approx \left(\frac{1}{2}\right)^{\frac{t}{27.8}} \).

Step 4: Take Logarithm of Both Sides

Take the logarithm of both sides to solve for \( \frac{t}{27.8} \): \( \log(0.2609) = \log\left(\left(\frac{1}{2}\right)^{\frac{t}{27.8}}\right) \). Apply the power rule of logarithms: \( \log(0.2609) = \frac{t}{27.8} \cdot \log\left(\frac{1}{2}\right) \).

Step 5: Solve for Time \( t \)

Solve for \( t \) by isolating it: \( t = 27.8 \cdot \frac{\log(0.2609)}{\log(0.5)} \). Calculate \( \log(0.2609) \approx -0.584 \) and \( \log(0.5) \approx -0.3010 \). Therefore, \( t \approx 27.8 \cdot \frac{-0.584}{-0.3010} \approx 53.9 \) days.

Key Concepts

Exponential Decay Formula

The exponential decay formula is a cornerstone of understanding how radioactive substances lose their mass over time. This formula is handy for many scientific fields, including physics and chemistry. In the context of radioactive decay, the formula is expressed as:\[N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}\]Here's what the variables mean:

• \( N(t) \) is the quantity of the radioactive substance remaining after time \( t \).
• \( N_0 \) is the initial quantity of the substance you started with.
• \( T_{1/2} \) represents the half-life of the substance, which is the time it takes for half of the radioactive substance to decay.
• \( t \) is the time elapsed, which we often aim to calculate.

The formula relies on the principle that the substance will reduce by half in a predictable, exponential manner at regular intervals given by its half-life. Thus, the use of \(\frac{1}{2}\) is because each half-life cuts the amount of substance remaining into two. Understanding this formula is crucial for solving real-world problems related to radioactive decay.

Half-Life Calculation

Calculating the half-life of a substance is essential for predicting how quickly a radioactive material will decay. The half-life is the period needed for half of the material to disappear. This concept is not only significant in science, but also in areas like archeology and medicine.For example, in the given problem, the half-life \( T_{1/2} \) of \( ^{51}\mathrm{Cr} \) is known to be 27.8 days. With this information and the exponential decay formula, we can determine how long it takes for a substance to reach a certain mass. By knowing the half-life:

• We can set up the equation with \( N_0 \) as the initial weight and \( N(t) \) as the desired final mass.
• Substitute \( T_{1/2} \) into the equation to find the time \( t \); this provides the decay duration for a particular mass reduction.

In our specific scenario of \( ^{51}\mathrm{Cr} \), the knowledge of the 27.8-day half-life is fundamental to calculating that it takes approximately 53.9 days for the sample to decay from 5.75 mg to 1.50 mg.

Logarithmic Operations

Logarithmic operations are mathematical tools that help us solve exponential equations by allowing us to deal with large-scale multiplicative processes in a more manageable additive manner. When dealing with radioactive decay, these operations let us solve for time variables like \( t \).In this context, once the exponential term is isolated, the natural next step is using logarithms to bring down the exponent:

• Taking the logarithm of both sides converts the exponential equation into a more straightforward linear form.
• Use the logarithmic identity \( \log(a^b) = b \cdot \log(a) \) to simplify further.

In our exercise, by applying the log to both sides, we reformulated the equation:\[\log(0.2609) = \frac{t}{27.8} \cdot \log\left(\frac{1}{2}\right)\]By solving the equation above — isolating \( t \) — and substituting the values, we reach the solution, determining that \( t \approx 53.9 \) days. This demonstrates the power of logarithms in simplifying and solving equations involving exponential decay.