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Chapter 42: Problem 49

Generally, more massive nuclides tend to be more unstable to alpha decay. For example, the most stable isotope of uranium, \({ }^{238} \mathrm{U},\) has an alpha decay half-life of \(4.5 \times 10^{9} \mathrm{y} .\) The most stable isotope of plutonium is \({ }^{244} \mathrm{Pu}\) with an \(8.0 \times 10^{7} \mathrm{y}\) half-life, and for curium we have \({ }^{248} \mathrm{Cm}\) and \(3.4 \times 10^{5} \mathrm{y}\). When half of an original sample of \({ }^{238} \mathrm{U}\) has decayed, what fraction of the original sample of (a) plutonium and (b) curium is left?

Short Answer

Expert verified
The fraction of plutonium left is approximately \(1.39 \times 10^{-17}\), and curium is essentially 0.

Step by step solution

01

Determine Uranium Half-life Count

First, note that half of the original sample of \({ }^{238} \text{U}\) will have decayed in one half-life period. This means the time elapsed is \(4.5 \times 10^9 \, \text{y}\).
02

Apply the Half-life Formula

The number of half-lives \( n \) for any nuclide over a time \( t \) can be calculated using the formula: \\[ n = \frac{t}{T_{1/2}} \]\where \(T_{1/2}\) is the half-life of the nuclide.
03

Calculate Remaining Plutonium

For plutonium \({ }^{244} \text{Pu}\), calculate the number of half-lives: \\[ n = \frac{4.5 \times 10^9 \, \text{y}}{8 \times 10^7 \, \text{y}} = 56.25 \]\The fraction remaining is \(\left(\frac{1}{2}\right)^n\): \\[ \left(\frac{1}{2}\right)^{56.25} \approx 1.39 \times 10^{-17} \]
04

Calculate Remaining Curium

For curium \({ }^{248} \text{Cm}\), calculate the number of half-lives: \\[ n = \frac{4.5 \times 10^9 \, \text{y}}{3.4 \times 10^5 \, \text{y}} = 13235.29 \]\The fraction remaining is \(\left(\frac{1}{2}\right)^n\): \\[ \left(\frac{1}{2}\right)^{13235.29} \approx 0 \]

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Key Concepts

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

Understanding Half-life
The concept of half-life is crucial in understanding radioactive decay processes. It refers to the time it takes for half of a radioactive sample to decay into another element or isotope. This is a constant property for each radioactive isotope.
To calculate the remaining fraction of a substance over time, the half-life formula is used:
  • \[ n = \frac{t}{T_{1/2}} \]
where \( t \) is the elapsed time and \( T_{1/2} \) is the half-life of the substance.
For example, if uranium-238 has a half-life of \(4.5 \times 10^9\) years, after that period, only half of the uranium would remain. This concept helps in estimating the age of substances and understanding their decay patterns.
Alpha Decay Explained
Alpha decay is a type of radioactive decay where an alpha particle is emitted from an atomic nucleus. This alpha particle consists of two protons and two neutrons, effectively a helium nucleus. When a nuclide undergoes alpha decay, it transforms into a new element with a mass number reduced by 4 and an atomic number reduced by 2.
Some key points about alpha decay include:
  • Alpha particles are relatively heavy and positively charged.
  • They have limited penetration ability, stopped by a sheet of paper or skin.
  • Common in heavy elements like uranium and plutonium.
Alpha decay contributes to the process by which unstable nuclides achieve stability by losing energy and mass, changing the elemental identity in the process.
Nuclides and Instability
Unstable nuclides are atoms with nuclei that have an imbalance of protons and neutrons, leading to instability. They tend to undergo radioactive decay to reach a more stable form. The instability in heavier elements like uranium, plutonium, and curium arises due to:
  • Large numbers of protons creating strong repulsive forces.
  • A mismatch in the proton to neutron ratio.
These forces and ratios contribute to the likelihood and type of decay these elements will undergo. Generally, larger isotopes are more unstable, making them more prone to forms of decay like alpha decay. This intrinsic instability propels the decay process as the element strives for a lower energy state.
The Role of Nuclear Physics
Nuclear physics plays a fundamental role in understanding the behaviors and interactions of atomic nuclei. It encompasses topics like radioactive decay, nuclear reactions, and applications such as nuclear power and medical imaging.
Key aspects of nuclear physics include:
  • Exploring the forces within the nucleus, including the strong nuclear force that holds the nucleus together.
  • Studying decay processes that reveal the mechanisms by which nuclei lose energy.
  • Understanding energy generation through nuclear fission and fusion.
Insights from nuclear physics not only illuminate the nature of unstable nuclides and decay processes like alpha decay but also have practical applications in various technologies and industries, impacting both energy production and healthcare advancements.

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Most popular questions from this chapter

A penny has a mass of 3.0 g. Calculate the energy that would be required to separate all the neutrons and protons in this coin from one another. For simplicity, assume that the penny is made entirely of \({ }^{63} \mathrm{Cu}\) atoms (of mass \(62.92960 \mathrm{u}\) ). The masses of the proton-plus-electron and the neutron are \(1.00783 \mathrm{u}\) and \(1.00866 \mathrm{u}\), respectively.

If the unit for atomic mass were defined so that the mass of \({ }^{1} \mathrm{H}\) were exactly \(1.000000 \mathrm{u},\) what would be the mass of (a) \({ }^{12} \mathrm{C}\) (actual mass \(12.000000 \mathrm{u}\) ) and (b) \({ }^{238} \mathrm{U}\) (actual mass \(238.050785 \mathrm{u}\) )?

The radionuclide \({ }^{56} \mathrm{Mn}\) has a half-life of \(2.58 \mathrm{~h}\) and is produced in a cyclotron by bombarding a manganese target with deuterons. The target contains only the stable manganese isotope \({ }^{55} \mathrm{Mn},\) and the manganese \(-\) deuteron reaction that produces \({ }^{56} \mathrm{Mn}\) is $${ }^{55} \mathrm{Mn}+\mathrm{d} \rightarrow{ }^{56} \mathrm{Mn}+\mathrm{p}$$ If the bombardment lasts much longer than the half-life of \({ }^{56} \mathrm{Mn}\), the activity of the \({ }^{56} \mathrm{Mn}\) produced in the target reaches a final value of \(8.88 \times 10^{10}\) Bq. (a) At what rate is \({ }^{56}\) Mn being produced? (b) How many \({ }^{56}\) Mn nuclei are then in the target? (c) What is their total mass?

The nuclide \({ }^{198} \mathrm{Au},\) with a half-life of \(2.70 \mathrm{~d},\) is used in cancer therapy. What mass of this nuclide is required to produce an activity of \(250 \mathrm{Ci} ?\)

The nuclide \({ }^{14} \mathrm{C}\) contains (a) how many protons and (b) how many neutrons?
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