/*! 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 28 Upon decaying \({ }^{234}\), Np ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 45: Problem 28

Upon decaying \({ }^{234}\), Np emits a beta particle. The residual heavy nucleus is also radioactive, and gives rise to \({ }^{2.5} \mathrm{U}\) by the radioactive process. What small particle is emitted simultaneously with the formation of uranium-235?

Short Answer

Expert verified
An alpha particle is emitted.

Step by step solution

01

Understanding the Problem

When neptunium (Np) decays, it emits a beta particle and changes into another element. We need to determine what small particle accompanies the formation of uranium-235 during this radioactive process.
02

Beta Decay Explanation

Beta decay occurs when a neutron in an atom's nucleus is transformed into a proton, an electron (which is the beta particle), and an antineutrino. The electron is what is typically considered the 'beta particle' emitted during decay.
03

Identifying the Initial and Final Elements

Neptunium-234 (symbolized as { }^{234}_{93}Np ext{) undergoes beta decay}. In beta decay, an element with atomic number increased by 1 is produced. Therefore, { }_{94} ext{Pu}, which is plutonium with a possible mass of 234, is formed first.
04

Understanding Subsequent Transformation

After the formation of plutonium-234, it undergoes further radioactive decay to form uranium-235. We need to determine the type of decay that turns plutonium-234 into uranium-235.
05

Determining the Second Decay Process

For the transformation from plutonium-234 (94 protons) to uranium-235 (92 protons and gain of neutrons), likely alpha decay occurs. This process emits an alpha particle (consisting of 2 protons and 2 neutrons) reducing the atomic number by 2, to form uranium-235.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Radioactive Decay
Radioactive decay is the natural process by which an unstable atomic nucleus loses energy by emitting radiation. This helps the nucleus reach a more stable state. There are multiple types of radioactive decay, each with its unique characteristics and emitted particles.

In the context of the given problem, neptunium-234 undergoes radioactive decay. Initially, it experiences beta decay, where a neutron transforms into a proton and releases a beta particle (electron) and an antineutrino. This results in the formation of plutonium-234. It's important to note that radioactive decay is vital for the stability of heavy elements. Without it, many atoms would remain unstable.

Later in the solution, after plutonium is formed, a second type of decay occurs: alpha decay. This results in the formation of uranium-235 from plutonium-234, further illustrating the chain of transformations radioactive elements can undergo as they attempt to stabilize themselves.
Neutron to Proton Conversion
The neutron to proton conversion is a crucial part of beta decay, a type of radioactive decay. Understanding this conversion helps explain how elements change into other elements during decay.

During beta decay, a neutron in the nucleus transforms into a proton. This transformation involves the emission of a beta particle, which is essentially a high-speed electron, and an antineutrino. The emission of these particles is what mediates the neutron-to-proton conversion. The addition of a proton increases the atomic number of the element by 1. Hence, neptunium-234, which has 93 protons, turns into plutonium-234 with 94 protons.

This process increases the stability of the nucleus as it steps closer to a stable proton-to-neutron ratio. The conversion is essential in understanding the small changes in the composition of the atom, that eventually lead to the formation of different elements.
Alpha Decay
Alpha decay is another significant form of radioactive decay, distinct from beta decay. It primarily occurs in heavy elements, allowing them to achieve stability by reducing their size.

In alpha decay, the nucleus ejects an alpha particle, which consists of 2 protons and 2 neutrons. This particle is the same as a helium nucleus. The emission of an alpha particle reduces the atomic number of the element by 2 and its mass number by 4.

For instance, in the exercise solution, plutonium-234 undergoes alpha decay to become uranium-235. This process involves losing 2 protons and 2 neutrons, explaining the transition from an element with 94 protons to one with 92 protons. The result is not only a new element but also a more stable nucleus.

Alpha decay plays a significant role in the nuclear decay series of heavy elements, leading them toward a more stable end state.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 \(1.00-\mathrm{g}\) sample is \(0.00100 \mathrm{~kg}\), which for radium of atomic number 226 is \((0.00100 / 226) \mathrm{kmol}\). 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 decay constant is $$ \lambda=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{(1620 \mathrm{y})\left(3.156 \times 10^{7} \mathrm{~s} / \mathrm{y}\right)}=1.36 \times 10^{-11} \mathrm{~s}^{-1} $$ Then \(\quad \frac{\Delta N}{\Delta t}=\lambda N=\left(1.36 \times 10^{-11} \mathrm{~s}^{-1}\right)\left(2.66 \times 10^{21}\right)=3.61 \times 10^{10} \mathrm{~s}^{-1}\) 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: $$ 1 \mathrm{Ci}=3.7 \times 10^{10} \text { disintegrations } / \mathrm{s} $$ 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.

How many protons, neutrons, and electrons are there in \((a){ }^{3} \mathrm{He},(b){ }^{12} \mathrm{C}\), and \((c)^{206} \mathrm{~Pb}\) ? (a) The atomic number of He is 2 ; therefore, the nucleus must contain 2 protons. Since the mass number of this isotope is 3 , the sum of the protons and neutrons in the nucleus must equal 3 ; therefore, there is 1 neutron. The number of electrons in the atom is the same as the atomic number, \(2 .\) (b) The atomic number of carbon is 6 ; hence, the nucleus must contain 6 protons. The number of neutrons in the nucleus is equal to \(12-6=6 .\) The number of electrons is the same as the atomic number, \(6 .\) (c) The atomic number of lead is 82 ; hence, there are 82 protons in the nucleus and 82 electrons in the atom. The number of neutrons is \(206-82=124\).

How much energy must a bombarding proton possess to cause the reaction \({ }^{7} \operatorname{Li}(p, n)^{7} \mathrm{Be}\) ? Give your answer to three significant figures. The reaction is as follows: $$ { }_{3}^{7} \mathrm{Li}+{ }_{1}^{1} \mathrm{H} \rightarrow{ }_{4}^{7} \mathrm{Be}+{ }_{0}^{1} n $$ where the symbols represent the nuclei of the atoms indicated. Because the masses listed in Table \(45-2\) include the masses of the atomic electrons, the appropriate number of electron masses \(\left(m_{e}\right)\) must be subtracted from the values given. Subtracting the total reactant mass from the total product mass gives the increase in mass as \(0.00176 \mathrm{u}\). (Notice that the electron masses cancel out. This happens frequently, but not always.) To create this mass in the reaction, energy must have been supplied to the reactants. The energy corresponding to \(0.00176 \mathrm{u}\) is \((931 \times 0.00176) \mathrm{MeV}=1.65 \mathrm{MeV}\). This energy is supplied as \(\mathrm{KE}\) of the bombarding proton. The incident proton must have more than this energy because the system must possess some KE even after the reaction, so that momentum is conserved. With momentum conservation taken into account, the minimum KE that the incident particle must have can be found with the formula $$ \left(1+\frac{m}{M}\right)(1.65) \mathrm{MeV} $$ where \(M\) is the mass of the target particle, and \(m\) that of the incident particle. Therefore, the incident particle must have an energy of at least $$ \left(1+\frac{1}{7}\right)(1.65) \mathrm{MeV}=1.89 \mathrm{MeV} $$

An experiment is done to determine the half-life of a radioactive substance that emits one beta particle for each decay process. Measurements show that an average of \(8.4\) beta particles are emitted each second by \(2.5 \mathrm{mg}\) of the substance. The atomic mass of the substance is 230 . Find the half-life of the substance.

Uranium- \(238\left({ }_{92}^{238} \mathrm{U}\right)\) is radioactive and decays into a succession of different elements. The following particles are emitted before the nucleus reaches a stable form: \(\alpha, \beta, \beta, \alpha, \alpha, \alpha, \alpha, \alpha, \beta, \beta, \alpha, \beta, \beta\), and \(\alpha\) ( \(\beta\) stands for "beta particle," \(e^{-}\) ). What is the final stable nucleus? The original nucleus emitted 8 alpha particles and 6 beta particles. When an alpha particle is emitted. \(Z\) decreases by 2, since the alpha particle carries away a charge of \(+2 e .\) A beta particle carries away a charge of \(-1 e\), and so as a result the charge on the nucleus must increase to \((Z+1) e\). We then have, for the final nucleus, $$ \begin{array}{l} \text { Final } Z=92+6-(2)(8)=82 \\ \text { Final } A=238-(6)(0)-(8)(4)=206 \end{array} $$ The final stable nucleus is \({ }_{82}^{206} \mathrm{~Pb}\).
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Fields in Physics

Read Explanation

Electricity

Read Explanation

Fluids

Read Explanation

Energy Physics

Read Explanation

Famous Physicists

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.