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Chapter 24: Problem 106

Exactly 0.1 of the radioactive nuclei in a sample decay per hour. Thus, after \(n\) hours, the fraction of nuclei remaining is \((0.900)^{n} .\) Find the value of \(n\) equal to one half-life.

Short Answer

Expert verified
The value of \(n\) equal to one half-life is approximately 6.6 hours.

Step by step solution

01

Understand the concept of half-life

Half-life is the time required for half of the radioactive nuclei in a sample to decay. This means that after one half-life, only half of the original radioactive nuclei will remain.
02

Set up the equation

We know that after one half-life, the fraction of nuclei remaining is 0.5. Therefore, we need to solve for the value of \(n\) in the equation \[(0.900)^n = 0.5\]
03

Take the natural logarithm of both sides

Apply the natural logarithm (ln) to both sides of the equation to make it easier to solve for \(n\):\[\text{ln}((0.900)^n) = \text{ln}(0.5)\]Using the power rule of logarithms, \(\text{ln}(a^b) = b \text{ln}(a)\), this becomes:\[n \text{ln}(0.900) = \text{ln}(0.5)\]
04

Solve for \(n\)

Isolate \(n\) by dividing both sides of the equation by \(\text{ln}(0.900)\):\[n = \frac{\text{ln}(0.5)}{\text{ln}(0.900)}\]Calculate the natural logarithms:\[\text{ln}(0.5) \text{ is approximately } -0.693\]\[\text{ln}(0.900) \text{ is approximately } -0.105\]Thus,\[n = \frac{-0.693}{-0.105} \text{ is approximately } 6.6\]

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Key Concepts

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

Radioactive decay
Radioactive decay is a natural process by which an unstable atomic nucleus loses energy. This happens by emitting radiation, such as alpha particles, beta particles, or gamma rays. Over time, the number of radioactive nuclei decreases. The speed at which a substance undergoes decay is characterized by its half-life.

The decay process is random for individual atoms, but predictable for a large number of atoms. In this specific exercise, we're looking at a scenario where exactly 0.1 of the radioactive nuclei decay per hour. This means that, with each passing hour, 10% of the nuclei decay, leaving 90% still radioactive.
Natural logarithm
The natural logarithm (ln) is a logarithm with the base of the constant e, approximately equal to 2.718. The natural logarithm is very useful in solving equations involving exponential decay, such as our radioactive decay problem.

When we take the natural logarithm of both sides of an equation, it allows us to simplify and solve equations where the variable is part of an exponent.

In our problem, to find the half-life, we need to solve the equation \( (0.900)^n = 0.5 \). Taking the natural logarithm of both sides gives us \( \text{ln}((0.900)^n) = \text{ln}(0.5) \). This step helps to bring down the exponent, making it easier to solve for n.
Exponential decay
Exponential decay refers to the process in which the quantity of a substance decreases at a rate proportional to its current value. This can be described by the formula \( N(t) = N_0 e^{-\lambda t} \), where:
\[ N(t) \] is the quantity at time t
\[ N_0 \] is the initial quantity
\[ \lambda \] is the decay constant
\[ t \] is time

In our context, after n hours, the fraction of nuclei remaining is \( (0.900)^n \). This indicates exponential decay where each hour, only 90% of the nuclei remain compared to the previous hour.
Logarithmic properties
Logarithms have several important properties that are particularly useful in simplifying and solving equations involving exponential functions. Key properties include:
  • The power rule: \( \text{ln}(a^b) = b \text{ln}(a) \)
  • The product rule: \( \text{ln}(xy) = \text{ln}(x) + \text{ln}(y) \)
  • The quotient rule: \( \text{ln}(\frac{x}{y}) = \text{ln}(x) - \text{ln}(y) \)

In our specific problem, we use the power rule to transform \( \text{ln}((0.900)^n) \) into \( n \text{ln}(0.900) \). This makes it easier to isolate n by dividing both sides of the equation by \( \text{ln}(0.900) \). Using these properties simplifies complex exponential equations, allowing us to solve for unknown variables efficiently.

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Most popular questions from this chapter

In what main way is fission different from radioactive decay? Are all fission events in a chain reaction identical? Explain.

Write balanced nuclear equations for the following: (a) Formation of \({ }^{186}\) Ir through electron capture (b) Formation of francium- 221 through \(\alpha\) decay (c) Formation of iodine- 129 through \(\beta^{-}\) decay

Arsenic- 84 decays with an energy of \(1.57 \times 10^{-15} \mathrm{~kJ}\) per nucleus. Convert this energy into (a) eV; (b) MeV.

The starship Voyager, like many other vessels of the newly designed \(24^{\text {th }}\) -century fleet, uses antimatter as fuel. (a) How much energy is released when \(1.00 \mathrm{~kg}\) each of antimatter and matter annihilate each other? (b) When the antimatter is atomic antihydrogen, a small amount of it is mixed with excess atomic hydrogen (gathered from interstellar space during flight). The annihilation releases so much heat that the remaining hydrogen nuclei fuse to form \({ }^{4} \mathrm{He} .\) If each hydrogen-antihydrogen collision releases enough heat to fuse \(1.00 \times 10^{5}\) hydrogen atoms, how much energy (in \(\mathrm{kJ}\) ) is released per kilogram of antihydrogen? (c) Which produces more energy per kilogram of antihydrogen, the procedure in part (a) or that in part (b)?

24.128 A metastable (excited) form of \({ }^{50} \mathrm{Sc}\) changes to its stable form by emitting \(\gamma\) radiation with a wavelength of \(8.73 \mathrm{pm}\). What is the change in mass of \(1 \mathrm{~mol}\) of the isotope when it undergoes this change?
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