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Chapter 38: Problem 12

Why are fission fragments necessarily radioactive?

Short Answer

Expert verified
Fission fragments are necessarily radioactive because they are typically produced in an unstable state with excess energy or neutrons. They undergo radioactive decay, emitting radiation to move towards a more stable state.

Step by step solution

01

Understanding Nuclear Fission

Nuclear fission is a reaction that occurs when the nucleus of an atom, typically a heavy one such as Uranium-235, is split into two smaller nuclei, known as fission fragments. When a neutron interacts with the nucleus of the Uranium atom, it gets absorbed, making the nucleus unstable. This results in the nucleus splitting into two smaller nuclei and a few spare neutrons.
02

Explain Radioactivity

Radioactivity refers to the process wherein unstable atomic nuclei lose energy (or decay) by emitting radiation. This process helps the nuclei to achieve a more stable form. The emitted radiation could be in the form of alpha, beta, or gamma radiation.
03

Relating Fission Fragments and Radioactivity

The two smaller nuclei, or fission fragments, produced from the nuclear fission process, are not typically stable. They have excess neutrons or energy that make them unstable. Therefore, these fission fragments undergo radioactive decay, emitting radiation in order to lose the excess energy or neutrons and reach a more stable state. Hence, fission fragments are necessarily radioactive.

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Key Concepts

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

Radioactivity
Radioactivity is a naturally occurring process. It occurs when unstable atomic nuclei release energy to become more stable. This process is fundamental to nuclear reactions.
Radioactive decay can involve the emission of:
  • Alpha particles: Consisting of 2 protons and 2 neutrons, similar to a helium nucleus.
  • Beta particles: High-energy electrons or positrons released from the nucleus.
  • Gamma rays: High-energy electromagnetic radiation emitted from the nucleus.
Each of these emissions helps the atomic nucleus to shed excess energy or particles, allowing it to move toward a stable configuration. Wherever radioactivity occurs, stability follows. This is crucial in nuclear fission, where newly formed fission fragments are inherently radioactive and decay to attain stability.
Uranium-235
Uranium-235 is a specific isotope of uranium known for its potency in nuclear reactions. It plays a central role in nuclear fission due to its unique properties.
Here are some key points about Uranium-235:
  • It's a heavy element with 92 protons and 143 neutrons in its nucleus.
  • U-235 is distinguished by its ability to sustain a nuclear chain reaction.
  • When struck by a neutron, it becomes unstable and splits, yielding energy, additional neutrons, and fission fragments.
This ability to sustain a reaction makes U-235 vital for producing energy in nuclear reactors and explosive energy in nuclear weapons. The process of splitting results in radioactivity, reinforcing why U-235 is pivotal in nuclear technology.
Fission Fragments
When a Uranium-235 nucleus undergoes fission, it splits into smaller parts called fission fragments. These fragments are critical components of the fission process, as they provide insights into nuclear stability and decay.
Fission fragments have characteristics such as:
  • Being typically unstable due to an imbalance in neutron-to-proton ratio.
  • Undergoing radioactive decay to reach a stable state.
  • Releasing additional neutrons that propagate the chain reaction.
The instability of these fragments makes them naturally radioactive. They emit radiation to correct imbalances in their nuclear composition, losing extra neutrons or energy until they achieve a stable form. Understanding fission fragments enhances our knowledge of both nuclear reactions and radioactivity.

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Most popular questions from this chapter

A family member is about to have a brain scan using technetium\(99^{*},\) an excited isotope with 6.01 -hour half-life. The hospital makes Tc-99* from the decay of molybdenum-99 \(\left(t_{1 / 2}=2.7 \text { days }\right)\) then delivers it to the nuclear medicine department. You're told that the \(\mathrm{Tc}-99^{*}\) will arrive 90 minutes after production, and that there must be 10 mg of it. The technician says she will produce \(12 \mathrm{mg}\) of \(\mathrm{Tc}-99^{* *} .\) Is that sufficient?

Nucleus A decays into B with decay constant \(\lambda_{\mathrm{A}}\) and B decays into a stable product \(C\) with decay constant \(\lambda_{B}\). A pure sample starts with \(N_{0}\) nuclei \(\mathrm{A}\) at \(t=0 .\) Find an expression for the total activity of the sample at time \(t.\)

How do (a) the number of nucleons and (b) the nuclear charge compare in the two nuclei \(_{17}^{35} \mathrm{Cl}\) and \(_{19}^{35} \mathrm{K} ?\)

Brachytherapy is a cancer treatment involving implantation of radioactive "seeds" at the tumor site. Iridium-192, often used for cancers of the head and neck, undergoes beta decay by electron capture with 74.2 -day half-life. Inner-shell electrons drop to the orbital occupied by the captured electron, resulting in emission of gamma rays that kill surrounding tumor cells. What percentage of initial Ir- 192 activity will remain one year after implant?

A buildup of fission products "poisons" a reactor, dropping the multiplication factor to \(0.992 .\) How long will it take the reactor power to decrease by half, assuming a generation time of \(0.10 \mathrm{s} ?\)
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