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Chapter 46: Problem 1

The binding energy per nucleon for \({ }^{238} \mathrm{U}\) is about \(7.6 \mathrm{MeV}\), while it is about \(8.6 \mathrm{MeV}\) for nuclei of half that mass. If a \({ }^{238} \mathrm{U}\) nucleus were to split into two equal-size nuclei, about how much energy would be released in the process? There are 238 nucleons involved. Each nucleon will release about \(8.6-7.6=1.0 \mathrm{MeV}\) of energy when the nucleus undergoes fission. The total energy liberated is therefore about \(238 \mathrm{MeV}\) or \(2.4 \times 10^{2} \mathrm{MeV}\).

Short Answer

Expert verified
The energy released is about 238 MeV.

Step by step solution

01

Understanding Binding Energy

Binding energy per nucleon is the average energy that binds each nucleon within the nucleus. The higher the binding energy per nucleon, the more stable the nucleus. In this problem, we are comparing the binding energies of a uranium nucleus before and after it splits.
02

Calculation of Energy Difference

Given that the binding energy per nucleon changes from 7.6 MeV to 8.6 MeV after the uranium nucleus splits into two equal parts, each nucleon releases energy. Calculate this difference: \(8.6 - 7.6 = 1.0\) MeV per nucleon.
03

Number of Nucleons Involved

A \(^{238} \text{U}\) nucleus consists of 238 nucleons. All of these nucleons are involved in the fission process, which results in the release of energy calculated per nucleon.
04

Calculation of Total Energy Released

Multiply the difference in energy per nucleon by the total number of nucleons to find the total energy released: \(238 \times 1.0 = 238\) MeV.

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Key Concepts

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

Binding Energy
Binding energy is a crucial concept in nuclear physics. It describes the energy that binds protons and neutrons together in the nucleus. Without this energy, the positively charged protons would repel each other due to their like charges. The binding energy per nucleon is an average measure of the energy that holds each nucleon in a nucleus. This value is critical because it relates to the stability of the nucleus. Nuclei with higher binding energy per nucleon are generally more stable and less likely to undergo spontaneous reactions like fission.

For instance, in uranium-238, the binding energy is about 7.6 MeV per nucleon. When the uranium nucleus splits into smaller nuclei, the binding energy per nucleon increases, leading to a release of energy. This energy difference plays a vital role in processes such as nuclear fission, providing a powerful source of energy.
Nuclear Fission
Nuclear fission is a process where a heavy nucleus, such as uranium-238, splits into two smaller and more stable nuclei. This reaction is a type of nuclear decay that releases a significant amount of energy. It occurs because the binding energy per nucleon of the resulting smaller nuclei is higher than that of the original uranium-238 nucleus.

During fission, the original nucleus absorbs a neutron and becomes unstable, subsequently breaking apart into two smaller nuclei and additional neutrons. This transformation releases energy, as seen from the increase in binding energy per nucleon. The released neutrons can, in turn, trigger more fission reactions, creating a chain reaction. This principle is the fundamental mechanism behind nuclear reactors and atomic bombs. Understanding fission and the conditions leading to it is crucial for harnessing nuclear energy safely.
Uranium-238
Uranium-238 is a naturally occurring isotope of uranium with a nucleus containing 92 protons and 146 neutrons, adding up to 238 nucleons in total. This isotope is relatively stable, with a half-life of about 4.5 billion years, making it one of the most abundant isotopes of uranium found in nature.

While uranium-238 itself is not highly radioactive, it plays a key role in nuclear reactions and the generation of energy. It serves as a "fertile" material in nuclear reactors because it can capture neutrons to eventually form plutonium-239, a "fissile" material. Though uranium-235 is typically used directly for fission in reactors, uranium-238's ability to transform into other usable isotopes makes it an essential part of nuclear technology.
  • Key properties: 238 nucleons, 92 protons
  • Half-life: 4.5 billion years
  • Role in nuclear reactors: Formation of plutonium-239
Energy Calculation
Energy calculation in nuclear physics often revolves around changes in binding energy during reactions. In the context of the original exercise, we deal with the energy release when uranium-238 undergoes fission. The approach to calculating this energy is straightforward, involving multiplying the change in binding energy per nucleon by the number of nucleons involved.

In the described scenario, the binding energy per nucleon increases from 7.6 MeV to 8.6 MeV after the fission. Thus, each nucleon contributes a 1.0 MeV increase of energy. With 238 nucleons in a uranium-238 nucleus, the total energy released during the fission process amounts to 238 MeV.

Understanding these calculations is essential to grasping how nuclear power is generated. This concept illustrates the immense potential energy that can be unleashed from nuclear reactions, emphasizing both the opportunities and risks of harnessing such power.

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Most popular questions from this chapter

The following fusion reaction takes place in the Sun and furnishes much of its energy: $$ 4{ }_{1}^{1} \mathrm{H} \rightarrow 4{ }_{2}^{4} \mathrm{He}+2_{+1}^{0} e+\text { energy } $$ where \(_{+1}^{0} e\) is a positron electron. How much energy is released as \(1.00 \mathrm{~kg}\) of hydrogen is consumed? The masses of \({ }^{1} \mathrm{H},{ }^{4} \mathrm{He}\), and \({ }_{+}{ }^{0} e\) are, respectively, \(1.007825,4.002604\), and \(0.000549 \mathrm{u}\), where atomic electrons are included in the first two values. Ignoring the electron binding energy, the mass of the reactants, 4 protons, is 4 times the atomic mass of hydrogen \(\left({ }^{1} \mathrm{H}\right)\), less the mass of 4 electrons: $$ \begin{aligned} \text { Reactant Mass } &=(4)(1.007825 \mathrm{u})-4 m_{e} \\ &=4.031300 \mathrm{u}-4 m_{e} \end{aligned} $$ where \(m_{e}\) is the mass of the electron (or positron). The reaction products have a combined mass $$ \begin{aligned} \text { Product mass } &=\left(\text { Mass of }_{2}^{4} \text { He nucleus }\right)+2 m_{e} \\ &=\left(4.002604 \mathrm{u}-2 m_{e}\right)+2 m_{e} \\ &=4.002604 \mathrm{u} \end{aligned} $$ The mass loss is therefore $$ (\text { Reactant mass })-(\text { Product mass })=\left(4.0313 \mathrm{u}-4 m_{e}\right)-4.0026 \mathrm{u} $$ Substituting \(m_{e}-0.000549\) u gives the mass loss as \(0.0265 \mathrm{u}\). But \(1.00 \mathrm{~kg}\) of \({ }^{1} \mathrm{H}\) contains \(6.02 \times 10^{26}\) atoms. For each four atoms that undergo fusion, \(0.0265 \mathrm{u}\) is lost. The mass lost when \(1.00 \mathrm{~kg}\) undergoes fusion is therefore $$ \begin{aligned} \text { Mass loss } / \mathrm{kg} &=(0.0265 \mathrm{u})\left(6.02 \times 10^{26} / 4\right)=3.99 \times 10^{24} \mathrm{u} \\ &=\left(3.99 \times 10^{24} \mathrm{u}\right)\left(1.66 \times 10^{-27} \mathrm{~kg} / \mathrm{u}\right)=0.00663 \mathrm{~kg} \end{aligned} $$ Then, from the Einstein relation. $$ \Delta \mathrm{E}=(\Delta m) \mathrm{c}^{2}=(0.00663 \mathrm{~kg})\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)^{2}=5.96 \times 10^{14} \mathrm{~J} $$

A tumor on a person's leg has a mass of \(3.0 \mathrm{~g} .\) What is the minimum activity a radiation source can have if it is to furnish a dose of \(10 \mathrm{~Gy}\) to the tumor in \(14 \mathrm{~min}\) ? Assume each disintegration within the source, on the average, provides an energy \(0.70 \mathrm{MeV}\) to the tumor. A dose of 10 Gy corresponds to \(10 \mathrm{~J}\) of radiation energy being deposited per kilogram. Since the tumor has a mass of \(0.0030 \mathrm{~kg}\), the energy required for a 10 Gy dose is \((0.0030 \mathrm{~kg})(10 \mathrm{~J} / \mathrm{kg})=0.030 \mathrm{~J}\). Each disintegration provides \(0.70 \mathrm{MeV}\), which in joules is $$ \left(0.70 \times 10^{6} \mathrm{eV}\right)\left(1.60 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)=1.12 \times 10^{-13} \mathrm{~J} $$ A dose of 10 Gy requires that an energy of \(0.030 \mathrm{~J}\) be delivered. That total energy divided by the energy per disintegration, yields the number of disintegrations: $$ \frac{0.030 \mathrm{~J}}{1.12 \times 10^{-13} \mathrm{~J} / \text { disintegration }}=2.68 \times 10^{11} \text { disintegrations } $$ They are to occur in \(14 \mathrm{~min}\) (or \(840 \mathrm{~s}\) ), and so the disintegration rate is $$ \frac{2.68 \times 10^{11}}{840 \mathrm{~s}} \text { disintegrations }=3.2 \times 10^{8} \text { disintegrations/s. } $$ Hence, the source activity must be at least \(3.2 \times 10^{8} \mathrm{~Bq}\). Since \(1 \mathrm{Ci}=3.70 \times 10^{10} \mathrm{~Bq}\), the source activity must be at least \(8.6 \mathrm{mCi}\).

An alpha-particle beam enters a charge collector and is measured to carry \(2.0 \times 10^{-14} \mathrm{C}\) of charge into the collector each second. The beam has a cross-sectional area of \(150 \mathrm{~mm}^{2}\), and it penetrates human skin to a depth of \(0.14 \mathrm{~mm}\). Each particle has an initial energy of \(4.0 \mathrm{MeV}\). The \(Q\) for such particles is about 15 . What effective dose, in \(\mathrm{Sv}\) and in rem, does a person's skin receive when exposed to this beam for \(20 \mathrm{~s}\) ? Take \(\rho=900 \mathrm{~kg} / \mathrm{m}^{3}\) for skin.

Lithium hydride, LiH, has been proposed as a possible nuclear fuel. The nuclei to be used and the reaction involved are as follows: $$ \begin{array}{ccc} { }_{3}^{6} \mathrm{Li} \\ 6.01513 & +{ }_{1}^{2} \mathrm{H} & & \rightarrow 2{ }_{2}^{4} \mathrm{He} \\ & 2.01410 & 4.00260 \end{array} $$ the listed masses being those of the neutral atoms. Calculate the expected power production, in megawatts, associated with the consumption of \(1.00 \mathrm{~g}\) of LiH per day. Assume 100 percent efficiency. Ignoring the electron binding energies, the change in mass for the reaction must be computed first: We find the loss in mass by subtracting the product mass from the reactant mass. In the process, the electron masses drop out and the mass loss is found to be \(0.02403 \mathrm{u}\). The fractional loss in mass is \(0.0240 / 8.029=2.99 \times 10^{-3}\). Therefore, when \(1.00 \mathrm{~g}\) reacts, the mass loss is $$ \left(2.99 \times 10^{-3}\right)\left(1.00 \times 10^{-3} \mathrm{~kg}\right)=2.99 \times 10^{-6} \mathrm{~kg} $$ This corresponds to an energy of $$ \Delta \mathrm{E}=(\Delta m) \mathrm{c}^{2}=\left(2.99 \times 10^{-6} \mathrm{~kg}\right)\left(2.998 \times 10^{\mathrm{s}} \mathrm{m} / \mathrm{s}\right)^{2}=2.687 \times 10^{11} \mathrm{~J} $$ Then $$ \text { Power }=\frac{\text { Energy }}{\text { Time }}=\frac{2.687 \times 10^{11} \mathrm{~J}}{86400 \mathrm{~s}}=3.11 \mathrm{MW} $$

When an atom of \({ }^{235} \mathrm{U}\) undergoes fission in a reactor, about \(200 \mathrm{MeV}\) of energy is liberated. Suppose that a reactor using uranium- 235 has an output of \(700 \mathrm{MW}\) and is 20 percent efficient. (a) How many uranium atoms does it consume in one day? ( \(b\) ) What mass of uranium does it consume each day? (a) Each fission yields $$ 200 \mathrm{MeV}=\left(200 \times 10^{6}\right)\left(1.6 \times 10^{-19}\right) \mathrm{J} $$ of energy. Only 20 percent of this is utilized efficiently, and so Usable energy per fission \(=\left(200 \times 10^{6}\right)\left(1.6 \times 10^{-19}\right)(0.20)=6.4 \times 10^{-12} \mathrm{~J}\) Because the reactor's usable output is \(700 \times 10^{6} \mathrm{~J} / \mathrm{s}\), the number of fissions required per second is $$ \text { Fissions/s }=\frac{7 \times 10^{8} \mathrm{~J} / \mathrm{s}}{6.4 \times 10^{-12} \mathrm{~J}}=1.1 \times 10^{20} \mathrm{~s}^{-1} $$ and Fissions/day \(=(86400 \mathrm{~s} / \mathrm{d})\left(1.1 \times 10^{20} \mathrm{~s}^{-1}\right)=9.5 \times 10^{24} \mathrm{~d}^{-1}\) (b) There are \(6.02 \times 10^{26}\) atoms in \(235 \mathrm{~kg}\) of uranium- \(235 .\) Therefore, the mass of uranium- 235 consumed in one day is $$ \text { Mass }=\left(\frac{9.5 \times 10^{24}}{6.02 \times 10^{26}}\right)(235 \mathrm{~kg})=3.7 \mathrm{~kg} $$
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