/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Plutonium-239 decays by alpha em... [FREE SOLUTION] | 91Ó°ÊÓ

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Plutonium-239 decays by alpha emission. Write out the equation from the process.

Short Answer

Expert verified
Plutonium-239 decays into Uranium-235 emitting an alpha particle.

Step by step solution

01

Understand Alpha Decay

Alpha decay occurs when a nucleus emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons, which is represented as the element Helium, with the symbol \(^4_2He\).
02

Identify Initial Nucleus

The exercise focuses on Plutonium-239 which has an atomic number of 94. This means the chemical symbol for Plutonium is \({}_{94}^{239}Pu\).
03

Determine Final Nucleus

When Plutonium-239 undergoes alpha decay, it loses 2 protons and 2 neutrons. Subtracting this from Plutonium's atomic and mass numbers: The new atomic number is \(94 - 2 = 92\) and the new mass number is \(239 - 4 = 235\).
04

Identify Resulting Element

The element with an atomic number of 92 is Uranium, which is represented as \( {}_{92}^{235}U \). Therefore, the resulting nucleus is Uranium-235.
05

Write the Nuclear Equation

The nuclear equation for the decay process is written as follows \[ {}_{94}^{239}Pu \rightarrow {}_{92}^{235}U + {}_{2}^{4}He \]. This equation illustrates Plutonium-239 decaying into Uranium-235 by emitting an alpha particle.

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Key Concepts

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

Nuclear Physics
Nuclear physics is the branch of physics that studies the components and behavior of atomic nuclei. At the heart of nuclear physics lies the understanding of nuclear interactions and radioactive processes. These processes include alpha decay, which is a form of radioactive decay where the atomic nucleus emits an alpha particle.
Alpha decay is particularly significant in nuclear physics because it changes the identity of the original element, transforming it into a different element altogether.
The study of nuclear physics involves investigating how these transformations happen and why they occur. By exploring nuclear physics, we can understand the mechanics of atoms and how they undergo changes on a nuclear level. This understanding provides insights into the stability of elements and isotopes, as well as the forces that hold nuclei together.
Radioactive Decay
Radioactive decay is the process through which unstable atomic nuclei lose energy by emitting radiation. In the case of alpha decay, an unstable nucleus releases an alpha particle, which is composed of 2 protons and 2 neutrons, effectively a helium nucleus.
This process reduces the atomic number of the original nucleus by 2, and its atomic mass number by 4. As a result, the original element transforms into a completely different element, a phenomenon integral to understanding the lifecycle of radioactive materials.
Radioactive decay is a spontaneous process determined by the instability of the nucleus, and it occurs naturally. For students studying nuclear physics, recognizing the types of decay is essential for solving related problems, such as determining the elements involved and predicting the type of radiation emitted.
  • Alpha Decay: Emission of an alpha particle (2 protons and 2 neutrons).
  • Beta Decay: Emission of a beta particle (electron or positron).
  • Gamma Decay: Emission of gamma radiation (high energy photons).
Nuclear Equation
A nuclear equation is a symbolic way to represent nuclear reactions and transformations. In the case of alpha decay, the nuclear equation illustrates how the atomic structure changes during the process.
For instance, in the decay of Plutonium-239, the nuclear equation is written as:\[{}_{94}^{239}Pu \rightarrow {}_{92}^{235}U + {}_{2}^{4}He\]
This equation shows the transformation from Plutonium-239 to Uranium-235 with the emission of an alpha particle.
The left side of the equation represents the initial element and its atomic and mass numbers, while the right side displays the new element formed and the particles emitted.
Understanding nuclear equations is vital for anyone studying nuclear physics because they depict the conservation of mass and charge. Through practice, students can learn to balance nuclear equations, ensuring the sum of atomic and mass numbers remains equal on both sides of the equation.

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Most popular questions from this chapter

How many protons, neutrons, and electrons does an atom of \({ }_{92}^{235} \mathrm{U}\) possess?

A certain isotope has a half-life of \(7.0 \mathrm{~h}\). How many seconds does it take for 10 percent of the sample to decay?

The half-life of radium is \(1.62 \times 10^{3}\) years. How many radium atoms decay in \(1.00 \mathrm{~s}\) in a \(1.00 \mathrm{~g}\) sample of radium? The atomic weight of radium is \(226 \mathrm{~kg} / \mathrm{kmol}\). A 1.00-g sample is \(0.00100 \mathrm{~kg}\), which for radium of atomic number 226 is \((0.00100 / 226)\) kmol. Since each kilomole contains \(6.02 \times 10^{26}\) atoms, $$ N=\left(\frac{0.00100}{226} \mathrm{kmol}\right)\left(6.02 \times 10^{26} \frac{\text { atoms }}{\mathrm{kmol}}\right)=2.66 \times 10^{21} \text { atoms } $$ The decar constanti Then $$ \begin{array}{l} \lambda=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{(1620 y)\left(3.156 \times 10^{7} \mathrm{~s} / \mathrm{y}\right)}=1.36 \times 10^{-11} \mathrm{~s}^{-1} \\ \frac{\Delta N}{\Delta t}=\lambda N=\left(1.36 \times 10^{-11} \mathrm{~s}^{-1}\right)\left(2.66 \times 10^{21}\right)=3.61 \times 10^{10} \mathrm{~s}^{-1} \end{array} $$ is the number of disintegrations per second in \(1.00 \mathrm{~g}\) of radium. The above result leads to the definition of the curie (Ci) as a unit of activity: \(1 \mathrm{Ci}=3.7 \times 10^{10}\) disintegrations/s \(/ \mathrm{s}\) Because of its convenient size, we shall sometimes use the curie in subsequent problems, even though the official SI unit of activity is the becquerel.

A 5.6-MeV alpha particle is shot directly at a uranium atom \((Z=\) 92). About how close will it get to the center of the uranium nucleus? At such high energies the alpha particle will easily penetrate the electron cloud and the effects of the atomic electrons can be ignored. We also assume the uranium atom to be so massive that it does not move appreciably. Then the original KE of the alpha particle will be changed into electrostatic potential energy. This energy, for a charge \(q^{\prime}\) at a distance \(r\) from a point charge \(q\), $$ \text { Potential energy }=q^{\prime} V=k_{0} \frac{q q^{\prime}}{r} $$ Equating the KE of the alpha particle to this potential energy, $$ \left(5.6 \times 10^{6} \mathrm{eV}\right)\left(1.60 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)=\left(8.99 \times 10^{9}\right) \frac{(2 e)(92 e)}{r} $$ where \(e=1.60 \times 10^{-19} \mathrm{C}\). We find from this that \(r=4.7 \times 10^{-14}\) \(\mathrm{m} .\)

Plutonium-239 decays as in the previous problem with a half-life of 24000 years. How much of an original quantity of plutonium will still exist 72000 years after it was produced in a reactor?

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