Problem 72 - University Physics with Modern Physics

Question

The nucleus \(^{15}_{8} \mathrm{O}\) has a half-life of 122.2 \(\mathrm{s}\); \(^{19}_{8} \mathrm{O}\) has a half-life of 26.9 s. If at some time a sample contains equal amounts of \(^{15}_{8} \mathrm{O}\) and \(^{19}_{8} \mathrm{O}\), what is the ratio of \(^{15}_{8} \mathrm{O}\) to \(^{19}_{8} \mathrm{O}\) (a) after 4.0 minutes and (b) after 15.0 minutes?

Step-by-Step Solution

Answer

The ratio is approximately 121.9 after 4 minutes and 685.4 after 15 minutes.

Step 1: Calculate Decay Constant

The decay constant \( \lambda \) is calculated using the formula \( \lambda = \frac{\ln(2)}{t_{1/2}} \), where \( t_{1/2} \) is the half-life. Calculate for both isotopes.For \( ^{15}_{8} \mathrm{O} \):\[ \lambda_{15} = \frac{\ln(2)}{122.2 \, \text{s}} \approx 0.00567 \, \text{s}^{-1} \]For \( ^{19}_{8} \mathrm{O} \):\[ \lambda_{19} = \frac{\ln(2)}{26.9 \, \text{s}} \approx 0.02578 \, \text{s}^{-1} \]

Step 2: Determine Number of Nuclei After Time

The number of nuclei remaining after time \( t \) can be calculated using \( N(t) = N_0 e^{-\lambda t} \). Given equal initial amounts \( N_0(15) = N_0(19) \), calculate \( N(t) \) for each isotope after 4.0 minutes and 15.0 minutes.

Step 3: Ratio After 4.0 Minutes

Convert 4.0 minutes to seconds: \( 4.0 \times 60 = 240 \) seconds.Apply the formula:For \( ^{15}_{8} \mathrm{O} \):\[ N(t)_{15} = N_0 e^{-0.00567 \times 240} \approx N_0 e^{-1.3608} \approx 0.256 N_0 \]For \( ^{19}_{8} \mathrm{O} \):\[ N(t)_{19} = N_0 e^{-0.02578 \times 240} \approx N_0 e^{-6.1872} \approx 0.0021 N_0 \]Calculate the ratio:\[ \text{Ratio}_{4 \text{ min}} = \frac{0.256 N_0}{0.0021 N_0} \approx 121.9 \]

Step 4: Ratio After 15.0 Minutes

Convert 15.0 minutes to seconds: \( 15.0 \times 60 = 900 \) seconds.Apply the formula:For \( ^{15}_{8} \mathrm{O} \):\[ N(t)_{15} = N_0 e^{-0.00567 \times 900} \approx N_0 e^{-5.103} \approx 0.0061 N_0 \]For \( ^{19}_{8} \mathrm{O} \):\[ N(t)_{19} = N_0 e^{-0.02578 \times 900} \approx N_0 e^{-23.202} \approx 0.0000089 N_0 \]Calculate the ratio:\[ \text{Ratio}_{15 \text{ min}} = \frac{0.0061 N_0}{0.0000089 N_0} \approx 685.4 \]

Key Concepts

Half-life calculation

The concept of half-life is crucial in understanding radioactive decay. Half-life is the amount of time it takes for half of the radioactive nuclei in a sample to decay. It's a measure of the stability of a radioactive isotope. A shorter half-life means the substance decays faster, while a longer half-life indicates slower decay.

To calculate the half-life, you don't actually need to wait for the nuclei to decay. Instead, you can use the known half-life value to understand how much of the substance remains after a certain time has passed. For example, in our problem, we were given the half-lives of isotopes: 122.2 seconds for \(^{15}_{8} \mathrm{O}\) and 26.9 seconds for \(^{19}_{8} \mathrm{O}\).

If the exercise asks you to determine the amount remaining after a time period like 4 or 15 minutes, you would use the half-life values to understand the decay process over these periods. This often involves using the decay formula and converting time into the same units as the given half-life, such as converting minutes into seconds to match the seconds in the half-life.

Decay constant

The decay constant, denoted by the Greek letter \( \lambda \), represents the probability of decay of a single radioactive atom per unit time. It's directly related to the half-life of a substance and provides a deeper insight into how quickly a radioactive material will decay.

• The formula: \( \lambda = \frac{\ln(2)}{t_{1/2}} \) connects half-life \( t_{1/2} \) and the decay constant \( \lambda \).


• The natural logarithm of 2 (\( \ln(2) \)) appears because it's the transition from half to one-third, a natural rate of exponential depletion.

For the given isotopes, using the formula, we calculated decay constants: approximately 0.00567 \, \text{s}^{-1} for \(^{15}_{8} \mathrm{O}\) and 0.02578 \, \text{s}^{-1} for \(^{19}_{8} \mathrm{O}\).

A larger decay constant indicates a quicker decay process, aligning with a shorter half-life. This means \(^{19}_{8} \mathrm{O}\) decays faster than \(^{15}_{8} \mathrm{O}\) due to its larger decay constant value.

Exponential decay

Exponential decay describes the process by which quantities decrease over time following an exponential function. This is typical in radioactive decay, where nuclear species lose particles at a rate proportional to their current amount. The mathematical expression for this is often written as: \( N(t) = N_0 e^{-\lambda t} \), where:

• \( N(t) \) is the number of radioactive atoms remaining after time \( t \).


• \( N_0 \) is the initial number of radioactive atoms.


• \( \lambda \) is the decay constant.


• \( t \) is time elapsed.

This formula shows that the quantity \( N(t) \) decays exponentially with time.

In the exercise, we applied this formula to determine how much of each isotope remains after set periods of 4 and 15 minutes. Noticeably, as the time doubles, the amount left doesn't just halve linearly: it follows the exponential curve specified by the decay constant.

Using this knowledge, after a long enough time, the more rapidly decaying \(^{19}_{8} \mathrm{O}\) reduces far quicker than \(^{15}_{8} \mathrm{O}\), changing the ratio of the isotopes significantly over time.