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A \(20.0\) kiloton atomic bomb releases as much energy as \(20.0\) kilotons of TNT \((1.0\) kiloton of TNT releases about \(5.0 \times 10^{12} \mathrm{~J}\) of energy). Recall that about \(2.0 \times 10^{2} \mathrm{MeV}\) of energy is released when each \({ }_{92}^{235} \mathrm{U}\) nucleus fissions. (a) How many \({ }_{92}^{235} \mathrm{U}\) nuclei are fissioned to produce the bomb's energy? (b) How many grams of uranium are fissioned? (c) What is the equivalent mass (in grams) of the bomb's energy?

Short Answer

Expert verified
(a) Approximately \(3.12 \times 10^{24}\) nuclei fissioned. (b) About \(1213\) grams of uranium fissioned. (c) The energy is equivalent to \(1.1\) grams.

Step by step solution

01

Calculate the total energy released by the bomb

The bomb releases energy equivalent to the energy from 20 kilotons of TNT. Since 1 kiloton of TNT releases \(5.0 \times 10^{12} \mathrm{~J}\), the total energy \(E\) is:\[E = 20.0 \times 5.0 \times 10^{12} = 1.0 \times 10^{14} \text{ J}\]
02

Convert MeV to Joules for uranium fission

We need to know the energy in joules when a single \({ }_{92}^{235} \mathrm{U}\) nucleus fissions. Given that each fission releases \(2.0 \times 10^{2} \text{ MeV}\), noting that \(1 \text{ MeV} = 1.602 \times 10^{-13} \text{ J}\), the energy per fission \(E_{fission}\) is:\[E_{fission} = 2.0 \times 10^{2} \times 1.602 \times 10^{-13} = 3.204 \times 10^{-11} \text{ J}\]
03

Calculate the number of fissioned nuclei

To find the number of fissioned nuclei \(N\), divide the total energy by the energy per fission:\[N = \frac{1.0 \times 10^{14}}{3.204 \times 10^{-11}} \approx 3.12 \times 10^{24} \]
04

Calculate the mass of fissioned uranium

The number of fissioned uranium atoms corresponds to a mass. Using Avogadro's number \(6.022 \times 10^{23}\) and the molar mass of uranium-235, which is about 235 g/mol, the mass \(m\) of fissioned uranium is:\[m = \frac{3.12 \times 10^{24}}{6.022 \times 10^{23}} \times 235 \approx 1213 \text{ grams}\]
05

Calculate the equivalent mass of the bomb's energy

To find the equivalent mass of the bomb's energy, use Einstein’s mass-energy equivalence \(E = mc^2\), where \(c = 3.0 \times 10^8 \text{ m/s}\) is the speed of light. Solving for mass \(m\):\[m = \frac{1.0 \times 10^{14}}{(3.0 \times 10^8)^2} \approx 1.1 \text{ grams}\]

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Key Concepts

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

Atomic Bomb Energy Calculation
To understand the energy calculation of an atomic bomb, we first need to know how much energy is released. A 20-kiloton atomic bomb releases energy equal to that of 20 kilotons of TNT. Since one kiloton of TNT releases approximately \(5.0 \times 10^{12}\, \text{J}\), we find the total energy released by multiplying:
  • Energy of 1 kiloton TNT: \(5.0 \times 10^{12}\, \text{J}\)
  • Total bomb energy: \(20 \times 5.0 \times 10^{12}\, \text{J} = 1.0 \times 10^{14}\, \text{J}\)
This calculation shows how much energy the atomic bomb releases in terms of joules. A clear understanding of this conversion helps grasp the immense power of nuclear explosions.
Uranium-235 Decay
Uranium-235 \((_{92}^{235}\text{U})\) is a radioactive isotope that undergoes nuclear fission. During this process, the nucleus of uranium splits into smaller parts, releasing a significant amount of energy. For every uranium-235 nucleus that undergoes fission, an energy release of \(2.0 \times 10^{2}\, \text{MeV}\) occurs.To use this information effectively, we convert this MeV energy into joules because calculations are often performed in SI units. Given:
  • 1 MeV = \(1.602 \times 10^{-13}\, \text{J}\)
  • Energy per fission: \(2.0 \times 10^{2} \times 1.602 \times 10^{-13}\, \text{J} = 3.204 \times 10^{-11}\, \text{J}\)
In this step, we align the unit of energy from nuclear decay with our calculation in joules, making further computations consistent and accurate.
Mass-Energy Equivalence
Mass-energy equivalence is a concept introduced by Einstein, linking mass \(m\) and energy \(E\) through the famous equation \(E = mc^2\). Here, \(c\) represents the speed of light, approximately \(3.0 \times 10^8\, \text{m/s}\). This principle explains how energy produced in nuclear reactions comes from a loss of mass:
  • Given total bomb energy \(E = 1.0 \times 10^{14}\, \text{J}\)
  • Equivalent mass: \(m = \frac{1.0 \times 10^{14}}{(3.0 \times 10^8)^2}\, \text{grams}\)
The actual mass converted to energy during a nuclear explosion is tiny but results in massive energy output. This crucial concept aids in understanding the power behind both nuclear energy and the destructive potential of nuclear weapons.
Energy Conversion in Physics
Energy conversion is a fundamental principle in physics, showing how energy transforms from one form to another. In nuclear physics, the transformation is from mass to energy. This principle is key to understanding nuclear fission's energy release. When uranium-235 nuclei undergo fission, the resultant energy is due to the conversion of a fraction of the nuclear binding mass into energy. This process converts a large amount of potential nuclear energy into kinetic energy and radiation, visible as heat and light in a nuclear explosion. Supporting points:
  • Nuclear reactions involve changing potential nuclear energy into usable forms.
  • Makes the magnitude of nuclear processes understandable when compared to traditional chemical reactions like burning fuel.
Comprehending this energy transformation assists students in connecting abstract physics concepts to real-world nuclear energy applications and their implications.

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Most popular questions from this chapter

co Concept Questions During a nuclear reaction, whether it is a fission reaction or a fusion reaction, energy is released. (a) Is the total mass of the particles after the reaction greater than, equal to, or less than the total mass of the particles before the reaction? (b) How is this difference in total mass related to the energy released by the reaction? Problem In one type of fusion reaction a proton fuses with a neutron to form a deuterium nucleus plus a \(\gamma\) -ray photon: \({ }_{1}^{1} \mathrm{H}+{ }_{0}^{1} \mathrm{n} \rightarrow{ }_{1}^{2} \mathrm{H}+\gamma .\) The masses are \({ }_{1}^{1} \mathrm{H}\) \((1.0078 \mathrm{u}),{ }_{0}^{1} \mathrm{n}(1.0087 \mathrm{u})\), and \({ }_{1}^{2} \mathrm{H}(2.0141 \mathrm{u})\). The \(\gamma\) -ray photon is massless. How much energy (in MeV) is released by this reaction?

A beam of particles is directed at a \(0.015-\mathrm{kg}\) tumor. There are \(1.6 \times 10^{10}\) particles per second reaching the tumor, and the energy of each particle is \(4.0 \mathrm{MeV}\). The \(\mathrm{RBE}\) for the radiation is 14 . Find the biologically equivalent dose given to the tumor in 25 \(\mathrm{s}\).

One kilogram of dry air at STP conditions is exposed to \(1.0 \mathrm{R}\) of \(\mathrm{X}\) -rays. One roentgen is defined by Equation \(32.1 .\) An equivalent definition can be based on the fact that an exposure of one roentgen deposits \(8.3 \times 10^{-3} \mathrm{~J}\) of energy per kilogram of dry air. Using the two definitions and assuming that all ions produced are singly charged, determine the average energy (in eV) needed to produce one ion in air.

Identify the unknown species \({ }_{Z}^{A} \mathrm{X}\) in the nuclear reaction \({ }_{11}^{22} \mathrm{Na}(d, \alpha){ }_{Z}^{A} \mathrm{X}\), where \(d\) stands for the deuterium isotope \({ }_{1}^{2} \mathrm{H}\) of hydrogen.

co Concept Questions (a) Which of the following are nucleons: protons, electrons, neutrons, \(\gamma\) -ray photons? (b) In a nuclear reaction, what is meant by the statement "The total number of nucleons is conserved"? (c) Which of the following have electric charge: protons, electrons, neutrons, \(\gamma\) -ray photons? (d) In a nuclear reaction, what is meant by the statement "The total electric charge is conserved"? Problem For each of the nuclear reactions listed below, determine the unknown particle \({ }_{Z}^{A} \mathrm{X}\) a. \(A_{Z} X+{ }_{7}^{14} N \rightarrow{ }_{1}^{1} H+{ }_{8}^{17} \mathrm{O}\) b. \({ }_{7}^{15} \mathrm{~N}+{ }_{Z}^{A} X \rightarrow{ }_{6}^{12} \mathrm{C}+{ }_{2}^{4} \mathrm{He}\) c. \({ }_{1}^{1} \mathrm{H}+{ }_{13}^{27} \mathrm{~A} 1 \rightarrow{ }_{Z}^{A} \mathrm{X}+{ }_{0}^{1} \mathrm{n}\) d. \({ }_{3}^{7} \mathrm{Li}+{ }_{1}^{1} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+{ }_{Z}^{A} X\)

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