Problem 71 - Master Resource Book in JEE Main Physics

Question

The half-life of radioactive Radon is \(3.8\) days. The time at the end of which ( \(1 / 20\) )th of the Radon sample will remain undecayed is (given \(\log _{10} e=0.4343\) ) (a) \(13.8\) days (b) \(16.5\) days (c) 33 days (d) 76 days

Step-by-Step Solution

Answer

The correct answer is (b) 16.5 days.

Step 1: Understand the Half-Life Concept

The half-life of a radioactive element is the time taken for half of the radioactive atoms in a sample to decay. For radon, the half-life is given as 3.8 days.

Step 2: Use the Formula for Decay Constant

The decay constant, \( \lambda \), is related to the half-life by the formula: \( T_{1/2} = \frac{0.693}{\lambda} \). Substitute \( T_{1/2} = 3.8 \) days to find \( \lambda \). Thus, \( \lambda = \frac{0.693}{3.8} \approx 0.182 \).

Step 3: Apply Exponential Decay Formula

The number of particles remaining after time \( t \) is given by \( N = N_0 e^{-\lambda t} \), where \( N_0 \) is the initial amount and \( N \) is the remaining amount \( \frac{1}{20}N_0 \). We set up the equation: \( \frac{1}{20}N_0 = N_0 e^{-\lambda t} \).

Step 4: Simplify the Equation

Cancel \( N_0 \) from both sides to get \( \frac{1}{20} = e^{-\lambda t} \). Take the natural logarithm on both sides to solve for \( t \):

Key Concepts

half-life calculation

The concept of half-life is essential in understanding radioactive decay. It's the time required for half of the radioactive atoms in a given sample to decay. This concept is particularly useful because it allows us to predict how a radioactive substance will decrease over time.

To calculate the half-life, we typically have a known quantity. For instance, with radon, its half-life is known to be 3.8 days. This means that every 3.8 days, half of any given quantity of radon will have decayed.

Why is Half-Life Important?

• It helps in understanding how quickly a substance will diminish.
• It is crucial for managing radioactive materials to ensure safety.
• It's widely applied in fields from archaeology (e.g., carbon dating) to nuclear medicine.

Calculating the half-life allows scientists and students to predict decay patterns and make educated estimates about the quantity of a substance in the future.

decay constant

The decay constant, denoted as \( \lambda \), is a parameter that reflects the rate of decay of a radioactive substance. It provides the exponential rate at which the substance will decrease over time.

Relationship with Half-Life
The decay constant is directly related to the half-life with the formula:\[ T_{1/2} = \frac{0.693}{\lambda} \]This formula stems from the natural logarithm value for 2, as half-life is essentially a repeated halving process. By rearranging the equation, we can solve for the decay constant if the half-life is known:

\[ \lambda = \frac{0.693}{T_{1/2}} \]

Example Calculation
Using radon as an example, with a half-life of 3.8 days:\[ \lambda = \frac{0.693}{3.8} \approx 0.182 \text{ days}^{-1} \]This value of \( \lambda \) indicates the probability per day that an individual radon atom will decay.

Significance of Decay Constant

• Helps in determining the rate of decay, which is critical for timing and dosage in medical treatments involving radioactive isotopes.
• Offers a universal way to compare decay rates of different substances.
• Used in complex calculations involving exponential decay equations.

exponential decay formula

The exponential decay formula is crucial for accurately determining how the quantity of a radioactive substance decreases over time.

The Formula
The general equation for exponential decay is:\[ N = N_0 e^{-\lambda t} \]Where:

• \( N_0 \) is the initial quantity of the substance.
• \( \lambda \) is the decay constant.
• \( t \) is time.
• \( N \) is the quantity remaining after time \( t \).

This formula helps us model the decline of radioactive materials efficiently.

Solving Problems
To use this formula, first calculate \( \lambda \) as mentioned before. Then, substitute known values into the equation to solve for the unknown, whether that's the remaining quantity \( N \), the initial quantity \( N_0 \), or the time \( t \).

Example of Application
For the given problem, where we desire \( \frac{1}{20} \, \)th of the initial amount to remain:1. Use the equation \( \frac{1}{20} N_0 = N_0 e^{-\lambda t} \).2. Cancel out \( N_0 \) from both sides to get \( \frac{1}{20} = e^{-\lambda t} \).3. Taking the natural logarithm:\[ \ln \left( \frac{1}{20} \right) = -\lambda t \]4. Solve for time \( t \).Understanding this formula enables precise estimations crucial for science and industry alike.