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Chapter 23: Problem 46

Americium-240 is made by bombarding plutonium-239 with \(\alpha\) particles. In addition to \(^{240} \mathrm{Am}\), the products are a proton and two neutrons. Write a balanced equation for this process.

Short Answer

Expert verified
The balanced nuclear equation is: \[^{239}_{94} \mathrm{Pu} + ^{4}_{2} \mathrm{He} \rightarrow ^{240}_{95} \mathrm{Am} + ^{1}_{1} \mathrm{p} + 2 \times ^{1}_{0} \mathrm{n}\]

Step by step solution

01

Identify Reactants and Products

Bombarding plutonium-239 with \( \alpha \) particles indicates that the reactants are plutonium-239 \((^{239}_{94} \mathrm{Pu})\) and an \( \alpha \) particle \((^{4}_{2} \mathrm{He})\). The products given are americium-240 \((^{240}_{95} \mathrm{Am})\), a proton \((^{1}_{1} \mathrm{p})\), and two neutrons \((2 \times ^{1}_{0} \mathrm{n})\).
02

Write the Nuclear Equation

Using the information from Step 1, the nuclear equation can be set up as follows:\[^{239}_{94} \mathrm{Pu} + ^{4}_{2} \mathrm{He} \rightarrow ^{240}_{95} \mathrm{Am} + ^{1}_{1} \mathrm{p} + 2 \times ^{1}_{0} \mathrm{n}\]
03

Verify Conservation of Mass Number

Check that the total mass number on both sides of the equation matches.- Reactants: \(239 + 4 = 243\)- Products: \(240 + 1 + 1 \times 2 = 243\)Both sides equal 243, so mass number is conserved.
04

Verify Conservation of Atomic Number

Ensure that the total atomic number on both sides of the equation is conserved.- Reactants: \(94 + 2 = 96\)- Products: \(95 + 1 + 0 \times 2 = 96\)Both sides equal 96, confirming conservation of atomic number.

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Key Concepts

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

Alpha Particle
An alpha particle is a type of nuclear emission that consists of 2 protons and 2 neutrons. It is essentially a helium nucleus. Alpha particles are released during certain types of radioactive decay and are considered high-energy particles.
They play a significant role in nuclear reactions, specifically in the formation of new elements when they interact with other atomic nuclei.
  • Alpha particles are positively charged because of the protons.
  • They are heavy and not very penetrating but can still cause significant damage to cells if inhaled or ingested.
  • In nuclear equations, an alpha particle is denoted as \( ^{4}_{2} \text{He} \) indicating its mass number of 4 and a charge of 2+
In the context of the given nuclear reaction, an alpha particle is used to bombard plutonium-239, leading to the formation of new nuclear products including americium-240.
Plutonium-239
Plutonium-239 (\(^{239}_{94}\text{Pu}\)) is a radioactive isotope of plutonium. It is well-known for its use in nuclear reactors and as a material for nuclear weapons, due to its fissile properties.
  • It has a half-life of around 24,100 years, making it relatively long-lived among radioactive elements.
  • Pu-239 is primarily used as fuel in nuclear power plants.
  • When Pu-239 is bombarded with an alpha particle, it can transform into another element like americium-240, illustrating its role in creating new isotopes.
In the reaction provided in the exercise, Pu-239 acts as one of the primary reactants, signifying its importance in the generation of americium-240.
Americium-240
Americium-240 (\(^{240}_{95}\text{Am}\)) is a radioactive isotope produced from nuclear reactions. It is not found naturally and is typically synthesized in laboratories or nuclear reactors.
  • Americium has several isotopes, with Am-240 being one of them due to the addition of an alpha particle to plutonium-239.
  • Although not as widely known or used as plutonium isotopes, Americanium isotopes have applications in smoke detectors and gauges.
  • The creation of Am-240 in a controlled nuclear reaction offers insight into how we can harness nuclear reactions for useful purposes.
In the context of the exercise, the formation of Am-240 highlights the transformative power of nuclear reactions, emphasizing their importance in modern chemistry.
Balanced Nuclear Equation
Balanced nuclear equations ensure the conservation of mass and atomic numbers. This principle is crucial in representing nuclear reactions accurately. To achieve this:
  • The sum of the mass numbers (top numbers) on both sides of the equation must be equal.
  • The sum of the atomic numbers (bottom numbers) must also be equal on both sides.
In our given reaction, Pu-239 and an alpha particle result in Am-240, a proton, and two neutrons.
The mass numbers add up to 243 on both the reactant and product sides \((239 + 4 = 240 + 1 + 2) \), confirming conservation. The atomic numbers also balance out to 96, maintaining the principle of conservation \((94 + 2 = 95 + 1 + 0 \times 2)\).Properly balancing nuclear equations is fundamental in studying nuclear chemistry and physics, helping to ensure the safe and effective management of nuclear materials.

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Most popular questions from this chapter

To synthesize the heavier transuranium elements, a nucleus must be bombarded with a relatively large particle. If you know the products are californium- 246 and four neutrons, with what particle would you bombard uranium-238 atoms?

A technique to date geological samples uses rubidium-\(87,\) a long-lived radioactive isotope of rubidium \(\left(t_{1 / 2}=\right.\) \(4.8 \times 10^{10}\) years). Rubidium-87 decays by \(\beta\) emission to strontium-87. If the rubidium-87 is part of a rock or mineral, then strontium-87 will remain trapped within the crystalline structure of the rock. The age of the rock dates back to the time when the rock solidified. Chemical analysis of the rock gives the amounts of \(^{87} \mathrm{Rb}\) and \(^{87}\) Sr. From these data, the fraction of \(^{87} \mathrm{Rb}\) that remains can be calculated. Analysis of a stony meteorite determined that \(1.8 \mathrm{mmol}\) of \(^{87} \mathrm{Rb}\) and \(1.6 \mathrm{mmol}\) of \(^{87} \mathrm{Sr}\) were present. Estimate the age of the meteorite. (Hint: The amount of \(^{87} \mathrm{Rb}\) at \(t_{0}\) is moles \(^{87} \mathrm{Rb}+\) moles \(^{87} \mathrm{Sr} .\))

Strontium-90 is a hazardous radioactive isotope that resulted from atmospheric nuclear testing. A sample of strontium carbonate containing \(^{90} \mathrm{Sr}\) is found to have an activity of \(1.0 \times 10^{3} \mathrm{dpm} .\) One year later the activity of this sample is 975 dpm. (a) Calculate the half-life of strontium-90 from this information. (b) How long will it take for the activity of this sample to drop to \(1.0 \%\) of the initial value?

(a) Which of the following nuclei decay by \(_{-1}^{0} \beta\) decay? $$^{1} \mathrm{H} \quad^{23} \mathrm{Mg} \quad^{32} \mathrm{P} \quad^{20} \mathrm{Ne}$$ (b) Which of the following nuclei decay by \(^{0}_{+1} \beta\) decay? $$^{235} \mathrm{U} \quad^{35} \mathrm{Cl} \quad^{38} \mathrm{K} \quad^{24} \mathrm{Na}$$

Describe how the concept of half-life for nuclear decay is used.
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