Problem 39 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Tritium has a half-life of \(12.26\) years. A \(5.0 \mathrm{ml}\) sample of triturated water has an activity of \(2.4 \times 10^{9} \mathrm{cpm} .\) How many years will it take for the activity to fall to \(3.0 \times 10^{8} \mathrm{cpm} ?\) (a) \(6.13\) (b) \(24.52\) (c) \(36.78\) (d) \(49.04\)

Step-by-Step Solution

Answer

It will take approximately 36.78 years for the activity to fall to \(3.0 \times 10^{8} \mathrm{cpm}\). The correct answer is (c) \(36.78\) years.

Step 1: Understanding Half-Life

Half-life is the amount of time it takes for half of a radioactive substance to decay. Tritium has a half-life of 12.26 years, which means every 12.26 years, its activity will reduce to half of its previous value.

Step 2: Calculating the Number of Half-Lives

Determine the number of half-lives required for the sample to decrease from an activity of \(2.4 \times 10^{9} \mathrm{cpm}\) to \(3.0 \times 10^{8} \mathrm{cpm}\). Use the formula \(N(t) = N_0 (\frac{1}{2})^{t/T_{1/2}}\), where \(N(t)\) is the remaining activity, \(N_0\) is the initial activity, \(T_{1/2}\) is the half-life, and \(t\) is the time elapsed. Here, you need to solve for \(t\).

Step 3: Solve for the Number of Half-lives

Using the formula, we have \(3.0 \times 10^{8} = 2.4 \times 10^{9} (\frac{1}{2})^{t/12.26}\). Taking the natural logarithm on both sides, we have \(\ \ln(3.0 \times 10^{8}) = \ln(2.4 \times 10^{9}) + (\frac{t}{12.26})\ln(\frac{1}{2})\).

Step 4: Calculate the Time Elapsed

Isolate \(t\) and solve for it using the logarithm properties and the values given to find the amount of time it takes for the activity to fall to the desired level.

Key Concepts

Half-Life Calculation

The half-life of a radioactive isotope is a central concept in understanding radioactive decay. It is defined as the time required for half of the radioactive nuclei in a sample to undergo decay. In our example, the isotope tritium has a half-life of 12.26 years, which means that after 12.26 years, only half of the initial amount of tritium would remain in an unaltered sample.

The mathematical approach to half-life problems involves understanding and applying the half-life formula. The formula is expressed as: \(N(t) = N_0 (\frac{1}{2})^{t/T_{1/2}}\), where \(N(t)\) is the quantity of the substance that still remains after time \(t\), \(N_0\) is the original quantity of the substance, and \(T_{1/2}\) represents the half-life of the substance.

When solving for time in half-life calculations, we often deal with sequential halving. In the example provided, the initial activity is 2.4 x 10^9 cpm, and we want to find out how long it will take for this activity to reduce to 0.3 x 10^9 cpm. By using the half-life formula, we can input our known values and solve for the time \(t\).

Radioactivity

Radioactivity is the process by which an unstable atomic nucleus loses energy by emitting energy in the form of particles or electromagnetic waves. This spontaneous emission is what we call radioactive decay, and it can involve the release of alpha particles, beta particles, gamma rays, and other particles.

In the context of our exercise, the substance in focus, tritium, is a radioactive isotope of hydrogen. The activity of a radioactive sample, such as our tritiated water, is measured in counts per minute (cpm) and reflects the number of decay events occurring per minute.

The rate of decay, and therefore the activity, will decrease over time as the atoms transform into a more stable state. It's crucial for students to grasp that the activity level directly correlates with the amount of radioactive material present: as the material decays, the activity decreases. In the exercise, the activity initially is 2.4 x 10^9 cpm, and understanding how this activity reduces over time is essential to solving the problem.

Exponential Decay

Radioactive decay follows a pattern known as exponential decay, which is a decreasing process that occurs at a rate proportional to the remaining amount of substance. What's important to note about exponential decay is that it is not linear; it doesn't decrease by the same amount over equal time intervals, but rather by proportionate fractions.

To illustrate, in the half-life calculations example, the quantity of a substance doesn't decline by a fixed amount each year but instead halves every 12.26 years. This relationship is why we employ the half-life formula and why the decay curve of radioactive substances is not straight but instead a curve that slopes downwards more steeply at the start and flattens out as time passes.

The concept of exponential decay is visually depicted on graphs using a curve that represents the decay over time. This visualization helps reinforce the understanding that as time goes by, the rate of decay lessens because there is less substance left to decay. Solving problems involving exponential decay requires comfortable handling of exponents and logarithms to isolate the variable of interest, which, in our problem, is the time \(t\) it takes for the activity to decrease to a certain level.