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Chapter 12: Problem 5

The half-life of \({ }^{24} \mathrm{Na}\) is \(15.0 \mathrm{~h}\). How long does it take for 80 percent of a sample of this nuclide to decay?

Short Answer

Expert verified
It takes approximately 34.8 hours for 80% of the sample of \(^{24}\mathrm{Na}\) to decay.

Step by step solution

01

Understand Half-life Concept

The half-life of a radioactive substance is the time required for half of the substance to decay. For \(^{24}\mathrm{Na}\), this half-life is given as \(15.0 \mathrm{~h}\). This means after 15 hours, half of the sample would have decayed.
02

Use Decay Formula

The decay of a radioactive substance can be calculated using the formula: \[ N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}} \]where \(N(t)\) is the remaining quantity of the substance at time \(t\), \(N_0\) is the initial quantity, and \(T_{1/2}\) is the half-life.
03

Calculate Time for 20% Remaining

We need to find the time \(t\) when 80% has decayed, meaning 20% is left. Let's set \(N(t) = 0.2 N_0\) in the decay formula:\[ 0.2 N_0 = N_0 \left(\frac{1}{2}\right)^{t/15} \]Solve for \(t\):\[ 0.2 = \left(\frac{1}{2}\right)^{t/15} \]
04

Solve for t Using Logarithms

Take the logarithm of both sides:\[ \log(0.2) = \log\left(\left(\frac{1}{2}\right)^{t/15}\right) \]Using the log power rule: \[ \log(0.2) = \frac{t}{15} \cdot \log\left(\frac{1}{2}\right) \]Rearranging for \(t\):\[ t = 15 \cdot \frac{\log(0.2)}{\log(0.5)} \]
05

Calculate the Value of t

Plug in the values:\[ t = 15 \cdot \frac{-0.69897}{-0.30103} \approx 15 \cdot 2.32189 \approx 34.83 \]Therefore, it takes approximately 34.83 hours for 80% of the sample to decay.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Half-Life
Half-life is a fundamental concept in understanding radioactive decay. It refers to the time required for half of a given amount of a radioactive substance to decay into a different substance. In simpler terms, if you start with a measurable amount of a radioactive material, half of it will have decayed after the material's half-life period.
For example, consider the isotope \(^{24} \text{Na}\) with a half-life of 15 hours. This means that if you start with 100 grams of \(^{24} \text{Na}\), after 15 hours, you will be left with 50 grams.
Half-life is an important concept because it provides a simple way to predict the decay of a radioactive substance over time without needing to analyze the full complexity of the decay process. Understanding half-life is essential for fields like archaeology, medicine, and nuclear physics, where accurate time estimations of decay can be crucial.
The Radioactive Decay Formula
The decay of radioactive substances can be predicted using the radioactive decay formula: \[ N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}} \]where:
  • \(N(t)\) is the quantity of substance remaining at time \(t\).
  • \(N_0\) is the original quantity of the substance.
  • \(T_{1/2}\) is the half-life of the substance.
This formula allows you to determine how much of a radioactive material will remain after a certain period.
If we want to know how long it will take for only 20% of \(^{24} \text{Na}\) to be left, we set \(N(t) = 0.2 N_0\) and solve for \(t\). This form is particularly useful because it simplifies the calculation process using the half-life measure.
Role of Logarithms in Solving Decay Problems
Logarithms are mathematical functions that help simplify the calculation of exponential decay processes like radioactive decay. When working with equations such as \[ 0.2 = \left(\frac{1}{2}\right)^{t/15} \]logarithms allow us to "bring down" the exponent, making the equation easier to solve.
Applying the power rule of logarithms, we take the logarithm of both sides: \[ \log(0.2) = \frac{t}{15} \cdot \log\left(\frac{1}{2}\right) \]This transformation lets us rearrange the equation to solve for \(t\): \[ t = 15 \times \frac{\log(0.2)}{\log(0.5)} \]
Using logarithms in decay questions helps in finding precise values for \(t\), especially when the decay percentage doesn't directly match a simple power of one-half. Understanding how to use logarithms can thus be a powerful tool in solving a variety of exponential decay problems efficiently.

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Most popular questions from this chapter

When an excited nucleus emits a gamma-ray pho?on, some of the excitation energy goes into the kinetic energy of the recoll of the nuclews. (a) Find the ratio between the recoil energy and the photon energy when the nucleus of an atom of mass \(200 \mathrm{u}\) emits a 2.0-MeV gamma ray- (b) The lifetime of an excited nuclear state is typically about \(10^{-14}\) s. Compare the corresponding uncertainty in the energy of the excited state with the recoil energy. (See Exercist 53 of Chap. 2 to leam how the Mossbauer effect can minimize nuclear recoll.)

What is the limitation on the fuel that can be used in a reactor whose moderator is ordinary water? Why is the situation different if the moderator is heavy water?

As discussed in this chapter, the heaviest nuclides are probably created in supernova explosions and become distributed in the galactic matter from which later stars (and their planets) form. Under the assumption that equal amounts of the \({ }^{235} \mathrm{U}\) and \({ }^{236} \mathrm{U}\) now in the earth were created in this way in the same supernova, calculate how long ago this occurred from their respective observed relative abundances of \(0.71\) and \(99.3\) percent and respective half-lives of \(7.0 \times 10^{8} \mathrm{y}\) and \(4.5 \times 10^{9} \mathrm{y}\).

Natural thorium consists entizely of the alpha-radioactive isotope \({ }^{232}\) Th which bas a half-life of \(1.4 \times 10^{10} \mathrm{y}\). If a rock sample known to have solidified \(3.5\) billion years ago contains \(0.100\) percent of \({ }^{232}\) Th today, what was the percentage of this nuclide it contained when the rock solidified?

The radionuclide \(230 \mathrm{U}\) decays into a lead isotope through the successive emissions of eight atpha particles and six electrons. Whal is the symbol of the lead isotope? What is the total energy released?
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