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

Question

How much time is required for a \(5.75-\mathrm{mg}\) 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 is approximately 53.9 days (option c).

Step 1: Understanding half-life and decay formula

The half-life of a substance is the time it takes for half of it to decay. The formula used to determine the amount of substance remaining after a certain period of time is given by \( N = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \), where \( N \) is the remaining amount, \( N_0 \) is the initial amount, \( T_{1/2} \) is the half-life, and \( t \) is the time elapsed.

Step 2: Set up the equation with known values

We know that the initial amount \( N_0 = 5.75 \text{ mg} \), the remaining amount \( N = 1.50 \text{ mg} \), and the half-life \( T_{1/2} = 27.8 \text{ days} \). Substituting these values into the decay formula gives us: \[ 1.50 = 5.75 \times \left( \frac{1}{2} \right)^{\frac{t}{27.8}} \].

Step 3: Solve the equation for \( t \)

First, divide both sides of the equation by 5.75 to find \( \left( \frac{1}{2} \right)^{\frac{t}{27.8}} \): \[ \frac{1.50}{5.75} = \left( \frac{1}{2} \right)^{\frac{t}{27.8}} \]. This simplifies to \[ 0.2609 \approx \left( \frac{1}{2} \right)^{\frac{t}{27.8}} \].

Step 4: Use logarithms to isolate \( t \)

Take the natural logarithm on both sides of the equation: \[ \ln(0.2609) = \ln\left(\left( \frac{1}{2} \right)^{\frac{t}{27.8}}\right) \]. Using the power rule of logarithms, this becomes \[ \ln(0.2609) = \frac{t}{27.8} \times \ln\left(\frac{1}{2}\right) \].

Step 5: Calculate \( t \)

Solve for \( t \) by isolating it: \[ t = \frac{27.8 \times \ln(0.2609)}{\ln\left(\frac{1}{2}\right)} \]. Calculate the values using a calculator. This yields \( t \approx 53.9 \) days.

Key Concepts

Half-Life

The concept of half-life is quite fascinating and is central to understanding radioactive decay. Essentially, the half-life of a radioactive substance is the time it takes for half of the substance to decay away. This means if you have a certain amount of a radioactive material, in one half-life, only half of it will remain. For example, if you start with 100 grams of a substance with a half-life of 1 year, you will have 50 grams left after 1 year, 25 grams after 2 years, and so on. Each period reduces the amount by half, hence the name "half-life."
This concept helps us determine how long it will take a material to decay to a specific other amount. By understanding half-life, we can predict and calculate the decay timeline of radioactive substances, which is crucial in fields like nuclear medicine, archaeology, and physics.

Decay Formula

To mathematically express how much of a radioactive substance remains after a certain time, we use the decay formula:
\[ N = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
This formula outlines a clear relationship between several key aspects:

• \(N\) is the amount of the substance left.
• \(N_0\) is the initial amount of the substance.
• \(t\) represents the time that has passed.
• \(T_{1/2}\) is the half-life of the substance.

The formula is based on the idea that a substance decays exponentially over time, where each half-life period reduces the substance by half of its present amount. Understanding this formula is key for calculating how much of a substance remains after a given time and is widely used in solving various scientific problems involving decay.

Exponential Decay

Radioactive decay is a classic example of exponential decay. This type of decay affects the amount of a substance as it decreases over time not by a set amount, but by a constant percentage. In radioactive decay, this constant is typically half of the existing amount after every half-life.
Exponential decay can be contrasted with linear decay, where a constant absolute amount is subtracted over time. In exponential decay, the equation integrates the process of continually reducing amounts, which is why the decay formula uses powers and roots to represent this naturally occurring process.
The exponential nature of decay means that even after many half-lives, a small amount of the original substance will always remain. Understanding exponential decay is vital in predicting how quickly a radioactive substance will reach a certain threshold, thus allowing us to solve problems related to time and decay.

Logarithms in Decay Calculations

Logarithms are powerful mathematical tools that help solve equations where a variable is an exponent, as seen in decay calculations. When we're dealing with exponential equations, logarithms can be used to extract that variable and make our calculations more manageable.
In the context of radioactive decay, once we set up the decay formula, we often need to determine the time elapsed (t) based on other known values. Transforming our exponential equation into a logarithmic form simplifies this process. By applying the natural logarithm, denoted as \( \ln \), we can effectively bring down the exponent.

• This transforms our earlier decay problem by allowing us to isolate (t) and solve for it directly.
• The process involves converting the expression, like \( \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \), using a logarithmic identity, \( \ln(a^b) = b \ln(a) \).

This powerful tool thus turns otherwise complicated calculations into straightforward algebraic processes, enabling us to solve for time in radioactive decay problems.