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Control rods in nuclear reactors tend to contain \({ }^{10} \mathrm{~B}\), which has a high neutron absorption cross section. \(^{81}\) What happens to this nucleus when it absorbs a neutron, and is the result stable? If not, track the decay chain until it lands on a stable nucleus.

Short Answer

Expert verified
Answer: The final stable nucleus is ${}^{11} \mathrm{~N}$.

Step by step solution

01

Write down the nuclear reaction after neutron absorption

When a \({ }^{10} \mathrm{~B}\) nucleus absorbs a neutron, we add the neutron to the existing nucleus to represent the new nucleus. In this case, it becomes \({ }^{11} \mathrm{~B}\). Hence, the initial nuclear reaction can be represented as follows: \({ }^{10} \mathrm{~B} + { }^1 \mathrm{n} \rightarrow { }^{11} \mathrm{~B}\)
02

Check the stability of the new nucleus

We need to verify whether the resulting \({ }^{11} \mathrm{~B}\) is a stable nucleus or not. Checking the stability, we find that \({ }^{11} \mathrm{~B}\) is actually unstable, undergoing beta decay.
03

Represent the beta decay for \({ }^{11} \mathrm{~B}\)

In beta decay, a neutron in the nucleus is converted to a proton, with the emission of an electron (more specifically, an electron plus an electron antineutrino). As a result, the atomic number increases, while the mass number remains the same. We write the beta decay for \({ }^{11} \mathrm{~B}\) as follows: \({ }^{11} \mathrm{~B} \rightarrow { }^{11} \mathrm{~C} + { }^{-1} \mathrm{e} + \bar{\nu}_\mathrm{e}\)
04

Check the stability of \({ }^{11} \mathrm{~C}\)

Now, the resulting nucleus is \({ }^{11} \mathrm{~C}\). We need to check if it is stable or not. If it is stable, we are done. But, in this case, we find that \({ }^{11} \mathrm{~C}\) is also unstable and undergoes beta decay as well.
05

Represent the beta decay for \({ }^{11} \mathrm{~C}\)

We repeat the process for beta decay. Writing the beta decay for \({ }^{11} \mathrm{~C}\), we get: \({ }^{11} \mathrm{~C} \rightarrow { }^{11} \mathrm{~N} + { }^{-1} \mathrm{e} + \bar{\nu}_\mathrm{e}\)
06

Check the stability of \({ }^{11} \mathrm{~N}\)

Now, we have reached \({ }^{11} \mathrm{~N}\), and we need to verify its stability. In this case, we find that \({ }^{11} \mathrm{~N}\) is a stable nucleus. Thus, we have reached a stable nucleus following the decay chain. In conclusion, the \({ }^{10} \mathrm{~B}\) nucleus undergoes the following decay chain after absorbing a neutron: \({ }^{10} \mathrm{~B} + { }^1 \mathrm{n} \rightarrow { }^{11} \mathrm{~B} \rightarrow { }^{11} \mathrm{~C} + { }^{-1} \mathrm{e} + \bar{\nu}_\mathrm{e} \rightarrow { }^{11} \mathrm{~N} + { }^{-1} \mathrm{e} + \bar{\nu}_\mathrm{e}\)

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Key Concepts

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

Beta Decay
Beta decay is a type of radioactive decay where a neutron in an unstable nucleus is transformed into a proton. This process emits an electron, also known as a beta particle, and an antineutrino. The atomic number of the nucleus increases by one, while the mass number remains unchanged. This is because a neutron is being converted into a proton without altering the total number of nucleons.
For example, when \({ }^{11} \mathrm{~B}\) undergoes beta decay, it turns into \({ }^{11} \mathrm{~C}\), as seen in the equation:
\[\]{ }^{11} \mathrm{~B} \rightarrow { }^{11} \mathrm{~C} + { }^{-1} \mathrm{e} + \bar{u}_\mathrm{e} \[\] In this reaction, the emission of an electron and an antineutrino balances the charge and lepton number. Beta decay helps in changing unstable nuclei into more stable forms, a crucial step in natural radioactive processes.
  • Neutron converts to proton.
  • Emission of electron and antineutrino.
  • Atomic number increases by one.
Neutron Absorption
Neutron absorption is a process where a neutron is captured by a nucleus, resulting in the formation of a new isotope. This process changes the identity of the nucleus by increasing its mass number by one.
In nuclear reactors, materials such as \({ }^{10} \mathrm{~B}\) are used to absorb excess neutrons and control the rate of the reaction. When \({ }^{10} \mathrm{~B}\) absorbs a neutron, it becomes \({ }^{11} \mathrm{~B}\):
\[\]{ }^{10} \mathrm{~B} + { }^1 \mathrm{n} \rightarrow { }^{11} \mathrm{~B}\[\] The capture of the neutron makes the new nucleus capable of undergoing subsequent nuclear reactions. Neutron absorption plays a key role in controlling chain reactions in nuclear plants as well as in various other nuclear applications.
  • Neutron capture forms a new isotope.
  • Mass number increases by one.
  • Used to control reactions in reactors.
Stable and Unstable Nuclei
Nuclei are considered stable if they do not undergo spontaneous radioactive decay. Stability in nuclei is often determined by the ratio of neutrons to protons. When this ratio is neither too high nor too low, the nucleus tends to be stable.
Unstable nuclei, on the other hand, tend to undergo radioactive decay to become more stable. For instance, \({ }^{11} \mathrm{~B}\) is unstable and will undergo beta decay to try and reach a stable state such as \({ }^{11} \mathrm{~N}\).
Understanding stability helps predict whether a nucleus will remain unchanged or undergo a series of transformations.
  • Stability depends on neutron-to-proton ratio.
  • Unstable nuclei seek stability through decay.
  • Stable nuclei do not change over time.
Nuclear Stability
Nuclear stability refers to a nucleus's ability to stay intact without breaking apart or transforming. Factors influencing stability include the neutron-to-proton ratio, binding energy, and nuclear forces within the nucleus.
Isotopes like \({ }^{11} \mathrm{~N}\) demonstrate nuclear stability because their internal forces are balanced, preventing spontaneous decay. Stable nuclei can be found on what is known as the "band of stability," a concept that helps illustrate which isotopes will likely exist without undergoing radioactive decay. During reactions, like nuclear fission, achieving nuclear stability ensures energy is released safely and efficiently.
  • Balance of nuclear forces determines stability.
  • Stable nuclei are on the band of stability.
  • Key to safe nuclear reactions.
Decay Chain Analysis
Decay chain analysis is the study of a sequence of radioactive decays that a nucleus undergoes to reach a stable form. This process involves mapping out each step a nucleus takes as it transforms through various intermediaries, often emitting particles in the process.
In the case of \({ }^{10} \mathrm{~B}\) absorbing a neutron, it undergoes several decay steps: \({ }^{11} \mathrm{~B}\) to \({ }^{11} \mathrm{~C}\), and finally to stable \({ }^{11} \mathrm{~N}\).
Each step in the decay chain is crucial to understanding both how an unstable nucleus seeks stability and the eventual endpoint of these transformations.
  • Tracks transformation to stable forms.
  • Includes intermediate decay steps.
  • Critical in predicting isotope behavior.

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Most popular questions from this chapter

Operating approximately 450 nuclear plants over about 60 years at a total thermal level of \(1 \mathrm{TW}\), we have had two major radioactive releases into the environment. If we went completely down the nuclear road and get all \(18 \mathrm{TW}^{89}\) this way, what rate of accidents might we expect, if the rate just scales with usage levels?

If a nuclear plant is built for \(\$ 10\) billion and operates for 50 years under an operating cost of \(\$ 100\) million per year, what is the cost to produce electricity, in \(\$ / \mathrm{kWh}\) assuming that the plant delivers power at a steady rate of \(1 \mathrm{GW}\) for the whole time?

On balance, considering the benefits and downsides of conventional nuclear fission, where do you come down in terms of support for either terminating, continuing, or expanding our use of this technology? Should we pursue breeder reactors at a large scale? Please justify your conclusion based on the things you consider to be most important.

Since each nuclear plant delivers \(\sim 1 \mathrm{GW}\) of electrical power, at \(\sim 40 \%\) thermodynamic efficiency this means a thermal generation rate of \(2.5\) GW. How many nuclear plants would we need to supply all 18 TW of our current energy demand? Since a typical lifetime is 50 years before decommissioning, how many days, on average would it be between new plants coming online (while old ones are retired) in a steady state?

A particular fission of \({ }^{235} \mathrm{U}+\mathrm{n}(\) total \(A=236)\) breaks up. One fragment has \(Z=54\) and \(N=86\), making it \({ }^{140} \mathrm{Xe} .\) If no extra neutrons are produced in this event, what must the other fragment be, so all numbers add up? Refer to a periodic table (e.g., Fig. B.1; p. 375 ) to learn which element has the corresponding \(Z\) value, and express the result in the notation \({ }^{\mathrm{A}} \mathrm{X}\).

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