Problem 93 - A Complete Resource Book in Chemistry for JEE Main

Question

The half-life of a radio-isotope is three hours. If the mass of the undecayed isotope at the end of 18 hours is \(3.125 \mathrm{~g}\), what was its mass initially? (a) \(300 \mathrm{~g}\) (b) \(200 \mathrm{~g}\) (c) \(180 \mathrm{~g}\) (d) \(400 \mathrm{~g}\)

Step-by-Step Solution

Answer

The initial mass of the isotope was 200 g (option b).

Step 1: Understand the Problem

We are asked to find the initial mass of a radio-isotope given its half-life and its final mass after a certain period. The half-life is 3 hours, and after 18 hours, the mass is 3.125 g. We need to find the initial mass from this information.

Step 2: Identify the Formula

The formula for the decay of a radioactive substance is given by the equation:\[ m(t) = m_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{\text{half}}}} \]where:- \( m(t) \) is the mass at time \( t \),- \( m_0 \) is the initial mass (what we need to find),- \( t \) is the total time elapsed,- \( t_{\text{half}} \) is the half-life.

Step 3: Plug in Known Values

Substitute the known values into the formula:- Final mass \( m(t) = 3.125 \text{ g} \)- Time \( t = 18 \text{ hours} \)- Half-life \( t_{\text{half}} = 3 \text{ hours} \)\[ 3.125 = m_0 \left( \frac{1}{2} \right)^{\frac{18}{3}} \]

Step 4: Simplify the Exponent

Calculate the exponent:\[ \frac{18}{3} = 6 \]Therefore, the equation becomes:\[ 3.125 = m_0 \left( \frac{1}{2} \right)^6 \]

Step 5: Solve for the Initial Mass

Calculate \( \left( \frac{1}{2} \right)^6 \):\[ \left( \frac{1}{2} \right)^6 = \frac{1}{64} \]Now the equation is:\[ 3.125 = m_0 \times \frac{1}{64} \]Multiply both sides by 64 to solve for \( m_0 \):\[ m_0 = 3.125 \times 64 = 200 \text{ g} \]

Step 6: Verify the Answer

Check the multiplication:\[ 3.125 \times 64 = 200 \]The calculations confirm that the correct initial mass was calculated.

Key Concepts

Half-life calculation

In radioactive decay, the half-life of a substance is the time it takes for half of the sample to decay. The concept revolves around the fact that every radioactive material decays over time, and this decay rate can be quantified using the half-life.

In this exercise, we know that the half-life of the isotope is 3 hours. This means that every 3 hours, half of the radioactive material transforms into a different element or a stable form. Therefore, if 18 hours have passed, you can determine how many half-lives have elapsed by dividing the total time by the half-life duration:

• Total time: 18 hours
• Half-life: 3 hours

Dividing the total time by the half-life gives:\[ \frac{18}{3} = 6 \]This tells us that 6 half-lives have passed in those 18 hours. This exponential reduction is critical in understanding how much of a radioactive substance remains after a given period.

Initial mass determination

Determining the initial mass of a radioactive substance is an essential part of understanding its decay. After identifying the number of half-lives that have elapsed, you can use the decay formula:
\[ m(t) = m_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{\text{half}}}} \]To find the initial mass \( m_0 \), we rearrange the equation using the known final mass \( m(t) \). Given that the final mass is \( 3.125 \, \text{g} \) after 18 hours, we need to backtrack to determine the original mass present before decay started:

• Final mass, \( m(t) = 3.125 \, \text{g} \)
• Number of half-lives, \( 6 \)

By calculating \( \left( \frac{1}{2} \right)^6 \) and setting \( m(t) = m_0 \times \left( \frac{1}{64} \right) \), we solve for the initial mass:\[ m_0 = 3.125 \times 64 = 200 \, \text{g} \]This computation shows that the original amount of the isotope was 200 grams before decay commenced.

Exponential decay formula

The exponential decay formula is fundamental to understanding radioactive materials' behaviour over time. It describes a process in which a quantity decreases at a rate proportional to its current value, a common trait in radioisotopes.

In mathematical terms, the decay is expressed as:\[ m(t) = m_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{\text{half}}}} \]Here:

• \( m(t) \) is the mass remaining after time \( t \).
• \( m_0 \) is the initial mass.
• \( t \) is the elapsed time.
• \( t_{\text{half}} \) is the half-life of the substance.

The fraction \( \left( \frac{1}{2} \right) \) reflects the halving of the mass with each passing half-life. As the time increases, the exponent \( \left( \frac{t}{t_{\text{half}}} \right) \) also increases, leading to a continuous degradation in the substance's mass. This formula also illustrates how exponentially smaller quantities remain as time progresses, highlighting the diminishing nature of radioactive decay.