/*! 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 61 Uranium-235 undergoes many diffe... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 61

Uranium-235 undergoes many different fission reactions. For one such reaction, when \(^{235} \mathrm{U}\) is struck with a neutron, \(^{144}\mathrm{Ce}\) and \(^{90}\mathrm{Sr}\) are produced along with some neutrons and electrons. How many neutrons and \(\beta\) -particles are produced in this fission reaction?

Short Answer

Expert verified
In the given fission reaction of Uranium-235, 2 neutrons and 4 尾-particles are produced along with Cerium-144 and Strontium-90.

Step by step solution

01

Determine the mass numbers and atomic numbers of the initial and final products

In this fission reaction, Uranium-235 is hit with a neutron, and the products are Cerium-144 and Strontium-90 along with some neutrons and beta particles. First, let's note down the mass numbers (A) and atomic numbers (Z) of each element involved in the reaction. - Uranium-235 (U): A = 235, Z = 92 - Neutron (n): A = 1, Z = 0 - Cerium-144 (Ce): A = 144, Z = 58 - Strontium-90 (Sr): A = 90, Z = 38 - Neutrons: A = 1, Z = 0 (each) - Beta particles (尾): A = 0, Z = -1 (each, since it is an electron emitted) We will denote the number of neutrons produced as x, and the number of beta particles produced as y.
02

Apply the conservation of mass number and atomic number

In order for the reaction to conserve mass number and atomic number, we need to satisfy the following equations: 1. Mass number conservation: 235 + 1 = 144 + 90 + x (U + n = Ce + Sr + neutrons) 2. Atomic number conservation: 92 = 58 + 38 + y (U = Ce + Sr + beta particles)
03

Solve the equations for x and y

Now, let's solve the equations to find x and y: 1. From the mass number conservation equation: 236 = 234 + x x = 2 (2 neutrons are produced) 2. From the atomic number conservation equation: 92 = 96 + y y = -4 (4 beta particles are produced) (Note that the negative sign indicates that 4 electrons are emitted as 尾-particles during the fission reaction.) #Conclusion# In this fission reaction of Uranium-235, 2 neutrons and 4 尾-particles are produced along with Cerium-144 and Strontium-90.

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.

Uranium-235
Uranium-235 is a naturally occurring isotope of uranium, which plays a crucial role in nuclear fission reactions. It has a mass number of 235 and an atomic number of 92. This means that its nucleus contains 235 nucleons (protons and neutrons combined) and 92 protons. What makes Uranium-235 particularly significant is its ability to sustain a chain reaction of fission, making it a valuable material for nuclear reactors and weapons.
  • Fissionable isotope: Can sustain chain reactions.
  • Mass number (A): 235 鈥� indicating nucleon count.
  • Atomic number (Z): 92 鈥� indicating the number of protons.
During a fission event, Uranium-235 absorbs a neutron and splits into two smaller nuclei, releasing energy and more neutrons, which can continue the reaction. Understanding its properties is essential in studying the dynamics of nuclear fission.
Neutron Interactions
Neutrons are key players in nuclear reactions and specifically in the fission of Uranium-235. Despite having a mass of 1 atomic mass unit (amu), neutrons carry no charge (Z=0), making them capable of easily penetrating atomic nuclei. In the context of Uranium-235, a neutron interacts by initiating the fission process once absorbed by the uranium nucleus.
  • Neutrons facilitate fission reactions.
  • A single neutron has a mass number (A) of 1 and atomic number (Z) of 0.
  • Neutron absorption by U-235 leads to its breakup into smaller nuclei.
These reactions not only result in smaller fission products but also release more free neutrons. These free neutrons can continue the chain reaction by interacting with other Uranium-235 nuclei, thus sustaining and amplifying the energy output.
Mass Number Conservation
Mass number conservation is a fundamental principle in nuclear reactions. It states that the total sum of mass numbers (A) before and after a reaction should remain unchanged. For the reaction of Uranium-235 with a neutron, we applied this principle to ensure all mass numbers total equally before and after fission occurred.
  • Initial mass numbers: Uranium-235 (235) + Neutron (1) = Total (236).
  • Products: Cerium-144 (144), Strontium-90 (90), and additional neutrons.
  • Conserved equation: 235 + 1 = 144 + 90 + x, solving for x gives the additional neutrons.
The conservation ensures no loss of mass occurs arbitrarily and helps us calculate the number of neutrons generated in the reaction, which is 2 in this case.
Atomic Number Conservation
The conservation of atomic numbers is another crucial principle in nuclear fission, signifying that the total number of protons is conserved. This conservation law helps to balance nuclear reactions and allows calculations of particles like beta particles (electrons emitted).
  • For Uranium-235: Initial atomic number is 92.
  • Products: Cerium-144 has 58 protons and Strontium-90 has 38 protons.
  • Conserved equation: 92 = 58 + 38 + y, solving for y reveals beta particles.
  • Beta particles have an atomic number (Z) of -1 for each emitted electron.
The result indicates the emission of 4 beta particles, ensuring that charge and proton counts are maintained from reactants to products in nuclear fission.

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

Radioactive cobalt-60 is used to study defects in vitamin \(\mathrm{B}_{12}\) absorption because cobalt is the metallic atom at the center of the vitamin \(\mathrm{B}_{12}\) molecule. The nuclear synthesis of this cobalt isotope involves a three-step process. The overall reaction is iron-58 reacting with two neutrons to produce cobalt-60 along with the emission of another particle. What particle is emitted in this nuclear synthesis? What is the binding energy in J per nucleon for the cobalt-60 nucleus (atomic masses: \(^{60} \mathrm{Co}=\) \(59.9338 \mathrm{u} ;^{-1} \mathrm{H}=1.00782\) u)? What is the de Broglie wavelength of the emitted particle if it has a velocity equal to \(0.90 c,\) where \(c\) is the speed of light?

Describe the relative penetrating powers of alpha, beta, and gamma radiation.

Given the following information: Mass of proton \(=1.00728 \mathrm{u}\) Mass of neutron \(=1.00866 \mathrm{u}\) Mass of electron \(=5.486 \times 10^{-4} \mathrm{u}\) Speed of light \(=2.9979 \times 10^{8} \mathrm{m} / \mathrm{s}\) Calculate the nuclear binding energy of \(_{12}^{24} \mathrm{Mg},\) which has an atomic mass of 23.9850 \(\mathrm{u}\) .

Phosphorus-32 2 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is 14.3 days. What mass of phosphorus- 32 is left from an original sample of 175 \(\mathrm{mg}\) \(\mathrm{Na}_{3}^{32} \mathrm{PO}_{4}\) after 35.0 days? Assume the atomic mass of \(^{32} \mathrm{P}\) is 32.0 \(\mathrm{u} .\)

Uranium-2355 undergoes a series of \(\alpha\) -particle and \(\beta\) -particle productions to end up as lead-207. How many \(\alpha\) particles and \(\beta\) particles are produced in the complete decay series?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

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.