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In a particular fission event in which \({ }^{235} \mathrm{U}\) is fissioned by slow neutrons, no neutron is emitted and one of the primary fission fragments is \({ }^{83}\) Ge. (a) What is the other fragment? The disintegration energy is \(Q=170 \mathrm{MeV}\). How much of this energy goes to (b) the \({ }^{83}\) Ge fragment and (c) the other fragment? Just after the fission, what is the speed of (d) the \({ }^{83}\) Ge fragment and (e) the other fragment?

Short Answer

Expert verified
Fragment is \(^{153}\text{Nd}\); \(63.8\ \text{MeV}\) to \(^{83}\text{Ge}\), \(106.2\ \text{MeV}\) to \(^{153}\text{Nd}\); calculate speeds f. both.

Step by step solution

01

Clarify the Fission Reaction

In nuclear fission of a \(^{235}\mathrm{U}\) atom by a neutron, the fission reaction can be written as \( \text{^{235}_{92}U} + \text{n} \rightarrow \text{^{83}_{32}Ge} + X \). Here, element \(X\) is the other fragment whose identity we need to determine.
02

Calculate the Other Fission Fragment

To find the other fragment \(X\), we use the conservation of nucleons:\[ 235 + 1 = 83 + A \]Solving gives \(A = 153\). For the atomic number:\[ 92 = 32 + Z \]Solving gives \(Z = 60\). The isotope is \(\text{^{153}_{60}Nd}\) (Neodymium).
03

Use Mass and Energy Definitions

The disintegration energy \(Q = 170\ \text{MeV}\) distributed to the fragments depends on their masses. Since energy and momentum are conserved, the kinetic energies are inversely proportional to their masses.
04

Calculate Energy Distribution

Using conservation of momentum \(m_1v_1 = m_2v_2\), where \(v\) represent fragment velocities and \(m\) their masses, we find kinetic energy:\[ KE = \frac{p^2}{2m} \]Thus energy ratio:\[ \frac{KE_1}{KE_2} = \frac{m_2}{m_1} \] is used to solve for individual energies.\(KE_{83\text{Ge}} = \frac{153}{83 + 153} Q \approx 63.8\ \text{MeV}\) and remaining for \(^{153}\text{Nd}\).
05

Calculate Velocities of Fragments

The speed of a fragment is determined from its kinetic energy:\[ KE = \frac{1}{2}mv^2 \]Apply to both elements using calculated energies to find speeds. For \(^{83}\text{Ge}\) and \(^{153}\text{Nd}\) fragments, speeds are .

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Key Concepts

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

Disintegration Energy
In nuclear fission, disintegration energy, often denoted by \( Q \), is the energy released when a heavy nucleus splits into smaller, lighter nuclei along with other particles. This energy comes from the binding energy difference between the original nucleus and the products of the fission process. During the fission of a \( ^{235}\text{U} \) atom, a \( Q \) value of 170 MeV is released. This energy is crucial because it is transformed into the kinetic energy of the fission fragments, making them move apart rapidly. Without this energy, the fission process would not lead to the separation of the newly formed nuclei. In practical applications, the disintegration energy is a central factor in determining the efficiency and the power output of nuclear reactors, as this is the energy that can be harnessed as usable power.
Conservation of Nucleons
Conservation of nucleons is a fundamental principle in nuclear reactions, including fission. It states that the total number of protons and neutrons (nucleons) remains constant before and after the reaction. This conservation is key to predicting the products of nuclear fission. In the exercise, the fission of \( ^{235}\text{U} \) by a slow neutron results in the formation of \( ^{83}\text{Ge} \) and another fragment. By using the conservation of nucleons, we add the nucleons of uranium and the neutron and set it equal to the sum of nucleons of \( ^{83}\text{Ge} \) and the unknown fragment. Solving the simple equations gives us the identity of the other fragment: \( ^{153}_{60}\text{Nd} \) (Neodymium), with a mass number of 153 and an atomic number of 60. Remember, this principle helps to ensure that our nuclear equations are balanced and respects fundamental laws of physics.
Neutron-Induced Fission
Neutron-induced fission occurs when a nucleus captures a neutron and subsequently undergoes fission, splitting into smaller nuclei. In this process, the nuclear binding energy is altered, releasing a significant amount of energy. For \( ^{235}\text{U} \), neutron absorption triggers fission, as the added neutron causes the nucleus to become unstable. The reaction \( ^{235}_{92}\text{U} + \text{n} \rightarrow ^{83}_{32}\text{Ge} + ^{153}_{60}\text{Nd} \) exemplifies neutron-induced fission. The neutron's role is pivotal, as it sets off the fission chain reaction used in both nuclear reactors and atomic bombs. While the fission of \( ^{235}\text{U} \) is particularly useful due to its ability to proceed with slow (thermal) neutrons, many other isotopes require fast neutrons. Understanding this aspect of neutron-induced fission is crucial for controlling nuclear reactions.
Kinetic Energy Distribution
The distribution of kinetic energy among fission fragments follows from the conservation of both energy and momentum. After fission, the disintegration energy \( Q \), becomes the kinetic energy of the resulting fragments. The energy distribution depends on the mass of the fragments: the lighter fragment will usually gain a larger share of the kinetic energy compared to a heavier one, due to their inverse mass relationship. For instance, in our example, \( ^{83}\text{Ge} \) and \( ^{153}\text{Nd} \) fragments share the 170 MeV disintegration energy proportionately to their masses.Mathematically, the kinetic energy \( KE \) of a fragment is determined by \( \frac{p^2}{2m} \), where \( p \) is its momentum. Using the steps from the exercise, we calculated that \( ^{83}\text{Ge} \) receives approximately 63.8 MeV. This understanding allows physicists and engineers to predict motion and energy distribution post-fission, key for reactor and bomb designs.

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Most popular questions from this chapter

(a) A neutron of mass \(m_{\mathrm{n}}\) and kinetic energy \(K\) makes a head-on elastic collision with a stationary atom of mass \(m\). Show that the fractional kinetic energy loss of the neutron is given by $$ \frac{\Delta K}{K}=\frac{4 m_{\mathrm{n}} m}{\left(m+m_{\mathrm{n}}\right)^{2}} $$ Find \(\Delta K / K\) for each of the following acting as the stationary atom: (b) hydrogen, (c) deuterium, (d) carbon, and (e) lead. (f) If \(K=1.00 \mathrm{MeV}\) initially, how many such head-on collisions would it take to reduce the neutron's kinetic energy to a thermal value \((0.025 \mathrm{eV})\) if the stationary atoms it collides with are deuterium, a commonly used moderator? (In actual moderators, most collisions are not head-on.)

Calculate the energy released in the fission reaction $$ { }^{235} \mathrm{U}+\mathrm{n} \rightarrow{ }^{141} \mathrm{Cs}+{ }^{93} \mathrm{Rb}+2 \mathrm{n} $$ Here are some atomic and particle masses. \(\begin{array}{cccc}{ }^{235} \mathrm{U} & 235.04392 \mathrm{u} & { }^{93} \mathrm{Rb} & 92.92157 \mathrm{u} \\\ { }^{141} \mathrm{Cs} & 140.91963 \mathrm{u} & \mathrm{n} & 1.00866 \mathrm{u}\end{array}\)

A star converts all its hydrogen to helium, achieving a \(100 \%\) helium composition. Next it converts the helium to carbon via the triple-alpha process, $$ { }^{4} \mathrm{He}+{ }^{4} \mathrm{He}+{ }^{4} \mathrm{He} \rightarrow{ }^{12} \mathrm{C}+7.27 \mathrm{MeV} $$ The mass of the star is \(4.6 \times 10^{32} \mathrm{~kg}\), and it generates energy at the rate of \(5.3 \times 10^{30} \mathrm{~W}\). How long will it take to convert all the helium to carbon at this rate?

Verify that the fusion of \(1.0 \mathrm{~kg}\) of deuterium by the reaction $$ { }^{2} \mathrm{H}+{ }^{2} \mathrm{H} \rightarrow{ }^{3} \mathrm{He}+\mathrm{n} \quad(Q=+3.27 \mathrm{MeV}) $$ could keep a \(100 \mathrm{~W}\) lamp burning for \(2.5 \times 10^{4} \mathrm{y}\).

The neutron generation time \(t_{\mathrm{gen}}\) in a reactor is the average time needed for a fast neutron emitted in one fission event to be slowed to thermal energies by the moderator and then initiate another fission event. Suppose the power output of a reactor at time \(t=0\) is \(P_{0}\). Show that the power output a time \(t\) later is \(P(t)\), where \(P(t)=P_{0} k^{t t \text { tgen }}\) and \(k\) is the multiplication factor. For constant power output, \(k=1\).

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