/*! 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 27 (II) Suppose that the average el... [FREE SOLUTION] | 91Ó°ÊÓ

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(II) Suppose that the average electric power consumption, day and night, in a typical house is \(880 \mathrm{~W}\). What initial mass of \({ }_{92}^{235} \mathrm{U}\) would have to undergo fission to supply the electrical needs of such a house for a year? (Assume \(200 \mathrm{MeV}\) is released per fission, as well as \(100 \%\) efficiency.)

Short Answer

Expert verified
The initial mass of \(^ {235}\text{U}\) needed is approximately 0.03374 kg.

Step by step solution

01

Calculate Total Energy Consumption in a Year

First, find the total energy consumption in one year for the house. The average power is given as 880 W. The total energy consumed in one year is calculated by multiplying the power by the total seconds in a year: \[ 880 \text{ W} \times 365 \times 24 \times 3600 \text{ s} = 2.77 \times 10^{10} \text{ J} \]
02

Convert Energy Per Fission to Joules

The energy released per fission is given as 200 MeV. Convert this energy to joules using the conversion factor: 1 MeV = \(1.602 \times 10^{-13}\) J. Therefore: \[ 200 \text{ MeV} = 200 \times 1.602 \times 10^{-13} \text{ J} = 3.204 \times 10^{-11} \text{ J} \]
03

Determine the Number of Fissions Required

Divide the total energy consumption by the energy released per fission to find the total number of fissions required: \[ \frac{2.77 \times 10^{10} \text{ J}}{3.204 \times 10^{-11} \text{ J/fission}} \approx 8.65 \times 10^{20} \text{ fissions} \]
04

Calculate Mass of Uranium-235 Required

Each fission uses one atom of \(^{235}\text{U}\). The mass of one atom of \(^{235}\text{U}\) can be calculated using Avogadro's number \(6.022 \times 10^{23}\) and the molar mass of \(^{235}\text{U}\) (235 g/mol): \[ \text{Mass of one } ^{235}\text{U atom} = \frac{235 \text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}} = 3.90 \times 10^{-22} \text{ g/atom} \] Multiply the mass of one atom by the number of fissions: \[ 8.65 \times 10^{20} \text{ atoms} \times 3.90 \times 10^{-22} \text{ g/atom} \approx 33.74 \text{ g} \]
05

Convert Mass to Kilograms

Convert the mass from grams to kilograms to provide the final result in a more standard unit of mass: \[ 33.74 \text{ g} = 0.03374 \text{ kg} \]

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Key Concepts

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

Uranium-235
Uranium-235, often abbreviated as U-235, is a naturally occurring isotope of uranium and is incredibly important in the nuclear physics realm due to its unique properties. It’s one of the isotopes used as fuel in nuclear reactors and weapons. What makes U-235 special is its ability to easily undergo nuclear fission, which is a process where the nucleus of an atom splits into smaller parts. This process releases a tremendous amount of energy.

One of the reasons U-235 is so valuable is its relatively low natural abundance, about 0.7% of uranium found on Earth is U-235. The rest is primarily U-238, which is non-fissile. To be used in reactors, uranium must be enriched to increase the concentration of U-235. Unlike U-238, U-235 can easily capture slow-moving neutrons and undergo fission, releasing energy and more neutrons that can perpetuate a chain reaction.

U-235's ability to sustain a chain reaction is the cornerstone of its application in power generation through nuclear fission. This unique capability makes it a critical component in the field of nuclear energy. Hence, understanding U-235 is essential for exploring the dynamics of nuclear reactors and the energy conversion within them.
Energy Conversion
In the context of nuclear physics, energy conversion is a key concept. When dealing with nuclear fission, like the process that occurs in Uranium-235, it refers to the transformation of nuclear energy into usable electrical energy. This is accomplished in nuclear power plants which utilize the energy released by the fission of U-235 to generate electricity.

To understand this conversion, visualizing the entire cycle is helpful:
  • When U-235 nuclei undergo fission, a tremendous amount of thermal energy is released.
  • This heat is used to produce steam from water.
  • The high-pressure steam then spins turbines.
  • The spinning turbines drive generators which convert mechanical energy into electrical energy.
This entire process has energy losses due to inefficiencies, but in theory, and as per the given textbook exercise, we assume a perfect conversion rate which is termed as 100% efficient. However, in real-world situations, efficiency varies but is a critical aspect of energy management in nuclear power systems. Understanding energy conversion is crucial for grasping how nuclear power plants contribute to electricity supply.
Nuclear Fission
Nuclear fission is a fundamental concept in nuclear physics, primarily involving the splitting of an atomic nucleus into smaller, lighter nuclei, accompanied by a few neutrons and a large amount of energy release. In our context, Uranium-235 is typically the isotope used as it readily undergoes fission upon absorbing a neutron.

The process of nuclear fission of U-235 involves several important steps:
  • A neutron collides with a U-235 nucleus.
  • This nucleus becomes unstable and splits into two smaller nuclei, which are called fission products.
  • Additional free neutrons are released, which can initiate further fission reactions, creating a self-sustaining chain reaction.
  • Significant amounts of energy, measured in Megaelectronvolts (MeV), are released during this process.
The energy released per fission event is substantial; in the exercise provided, it's noted to be 200 MeV. This energy is what we harness in nuclear reactors to produce electricity.

The chain reaction capability makes U-235 valuable not just for energy production but also highlights safety and control challenges in nuclear reactors. Understanding nuclear fission allows us to comprehend the mechanisms of energy generation and the underlying physics that govern the behavior of nuclear reactors.

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Most popular questions from this chapter

(I) Is the reaction \(\mathrm{n}+238 \mathrm{U} \rightarrow^{239} \mathrm{U}+\gamma\) possible with slow neutrons? Explain.

(a) What mass of \({ }_{92}^{235} \mathrm{U}\) was actually fissioned in the first atomic bomb, whose energy was the equivalent of about 20 kilotons of TNT (1 kiloton of TNT releases \(5 \times 10^{12} \mathrm{~J}\) )? (b) What was the actual mass transformed to energy?

(I) What is the effective cross section for the collision of two hard spheres of radius \(R_{1}\) and \(R_{2} ?\)

The early scattering experiments performed around 1910 in Ernest Rutherford's laboratory in England produced the first evidence that an atom consists of a heavy nucleus surrounded by electrons. In one of these experiments, \(\alpha\) particles struck a gold-foil target \(4.0 \times 10^{-5} \mathrm{~cm}\) thick in which there were \(5.9 \times 10^{28}\) gold atoms per cubic meter. Although most \(\alpha\) particles either passed straight through the foil or were scattered at small angles, approximately \(1.6 \times 10^{-3}\) percent were scattered at angles greater than \(90^{\circ}\) -that is, in the backward direction. ( \(a\) ) Calculate the cross section, in barns, for backward scattering. ( \(b\) ) Rutherford concluded that such backward scattering could occur only if an atom consisted of a very tiny, massive, and positively charged nucleus with electrons orbiting some distance away. Assuming that backward scattering occurs for nearly direct collisions (i.e., \(\sigma \approx\) area of nucleus), estimate the diameter of a gold nucleus.

(II) Huge amounts of radioactive 131 \(\mathrm{I}\) were released in the accident at Chernobyl in \(1986 .\) Chemically, iodine goes to the human thyroid. (Doctors can use it for diagnosis and treatment of thyroid problems.) In a normal thyroid, \(\frac{131}{53} \mathrm{I}\) absorption can cause damage to the thyroid. (a) Write down the reaction for the decay of \(\frac{131}{53} \mathrm{I}\) (b) Its half-life is 8.0 \(\mathrm{d}\) ; how long would it take for ingested 131 \(\mathrm{I}\) to become 7.0\(\%\) of the initial value? (c) Absorbing 1 \(\mathrm{mCi}\) of \(\frac{131}{53} \mathrm{I}\) can be harmful; what mass of iodine is this?

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