Problem 35 - Master Resource Book in JEE Main Physics

Question

The radioactivity of a given sample of whisky due to tritium (half-life \(12.3 \mathrm{yr}\) ) was found to be only \(3 \%\) of that measured in a recently purchased bottle marked 7 years old. The sample must have been prepared about (a) 220 years back (b) 300 years back (c) 400 years back (d) 70 years back

Step-by-Step Solution

Answer

The sample must have been prepared about 70 years ago (Option d).

Step 1: Understanding Half-life Concept

The half-life of a substance is the time it takes for half of the radioactive particles to decay. The half-life of tritium is given as 12.3 years. This means that every 12.3 years, the remaining amount of tritium will be half of its previous amount.

Step 2: Identifying Initial and Remaining Amounts

Initially, the bottle had 100% radioactivity. The current sample shows 3% of the original radioactivity. This means the sample's tritium has decayed to 3% of its initial amount.

Step 3: Setting Up the Decay Formula

We use the formula for radioactive decay: \[ R = R_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]where \(R\) is the remaining radioactivity (3), \(R_0\) is the initial radioactivity (100), \(t\) is the time elapsed, and \(T_{1/2}\) is the half-life (12.3 years).

Step 4: Solving the Decay Formula

Plug in the known values into the equation:\[ 3 = 100 \left( \frac{1}{2} \right)^{\frac{t}{12.3}} \]Divide both sides by 100:\[ 0.03 = \left( \frac{1}{2} \right)^{\frac{t}{12.3}} \]

Step 5: Using Logarithms to Solve for Time

Take the logarithm base 10 of both sides:\[ \log_{10} 0.03 = \frac{t}{12.3} \cdot \log_{10} \left( \frac{1}{2} \right) \]Solving for \(t\):\[ t = \frac{\log_{10} 0.03}{\log_{10}(0.5)} \times 12.3 \]

Step 6: Calculating Time Elapsed

Calculate the value: \[ \log_{10} 0.03 \approx -1.5229, \quad \log_{10}(0.5) \approx -0.3010 \]\[ t \approx \frac{-1.5229}{-0.3010} \times 12.3 \approx 62.2 \text{ years} \]

Step 7: Determining Total Time Since Preparation

The whisky was originally marked as 7 years old. Thus, 7 years ago, the preparation time was:\[ t + 7 \approx 62.2 + 7 = 69.2 \text{ years} \]

Key Concepts

Radioactive Decay

Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. In essence, it transforms into a different state or another element. This process occurs naturally and is characterized by a decrease in the number of radioactive atoms over time. The rate of decay is exponential, meaning it decreases rapidly at first and slows down over time.

One of the key aspects of radioactive decay is the half-life. The half-life is the time required for half of the radioactive nuclei in a sample to decay. For tritium, a radioactive isotope of hydrogen, this half-life is 12.3 years.

Understanding the concept of half-life is crucial in calculating the age of samples by measuring how much of the radioactive element remains. As seen in our exercise, the half-life concept helps determine how long ago a particular sample, like whisky, was bottled. This is done by observing the percentage of original radioactivity that remains and comparing it to the current state of decay.

Tritium

Tritium ( ^3 H) is a radioactive isotope of hydrogen containing one proton and two neutrons. It is naturally occurring and can also be produced in nuclear reactors. Though it is not dangerous to handle in small amounts, it does emit low-energy beta particles.

Its applications are diverse, from luminescent paints to a tracer in biochemical research. Its primary medical and industrial utility derives from its relatively short half-life, making it useful for dating and labeling studies. In the context of our exercise, tritium's half-life of 12.3 years allows calculating the age of various samples via the decay formula.

Tritium is found in trace amounts in the water and is particularly interesting in studies of environmental science and climatology. In conclusion, understanding tritium's decay properties allows scientists to harness its potential in applications where a defined isotopic age is pivotal.

Logarithmic Calculations

Logarithmic calculations are essential when dealing with exponential functions, such as those found in radioactive decay. A logarithm is the inverse operation to exponentiation, meaning it helps to solve equations where the unknown appears as an exponent.

When calculating the time elapse since the preparation of a sample, logarithms enable us to rearrange the decay formula. For instance, in our exercise, the equation was simplified by taking the logarithm of both sides. This conversion is crucial as it provides a means to solve for time (\(t\)), by isolating it in the logarithmic equation:

\[ t = \frac{\log_{10} 0.03}{\log_{10}(0.5)} \times 12.3 \]

In this equation, the logarithm of the remaining percentage compared to the original, divided by the logarithm of a half (from the decay process), provides the multiple of half-lives that have passed. When computed, this shows exactly how much time has elapsed since the sample was first prepared. This technique is incredibly powerful and widely used in scientific calculations where exponential growth or decay is involved.