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Which of the following decay modes is due to a transition between states of the same nucleus? a) alpha decay c) gamma decay b) beta decay d) none of the above

Short Answer

Expert verified
Answer: Gamma decay.

Step by step solution

01

Analyzing alpha decay

In alpha decay, an unstable nucleus emits an alpha particle (which consists of 2 protons and 2 neutrons). This process leads to the formation of a new nucleus, with a mass number decreased by 4 and an atomic number decreased by 2. Since a new nucleus is formed, alpha decay does not involve a transition between states of the same nucleus.
02

Analyzing gamma decay

In gamma decay, a nucleus transitions from a higher energy state to a lower energy state by emitting a gamma photon. The nucleus remains the same in terms of its mass number and atomic number, it merely transitions between energy states. Therefore, gamma decay involves a transition between states of the same nucleus.
03

Analyzing beta decay

In beta decay, a neutron in the nucleus is converted into a proton (when a beta-minus particle is emitted) or vice versa (when a beta-plus particle is emitted). In both cases, this causes the atomic number to change, and thus, a new nucleus is formed. Therefore, beta decay does not involve a transition between states of the same nucleus.
04

Concluding the answer

Based on the analysis of the given decay modes, we can conclude that gamma decay (option c) is the only decay mode that involves a transition between states of the same nucleus.

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Key Concepts

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

Alpha Decay
When we delve into the world of nuclear physics, we encounter a phenomenon known as alpha decay. This process involves an unstable atomic nucleus that releases an alpha particle, a chunk of matter consisting of two protons and two neutrons—the same configuration as a helium atom's nucleus.

Let's imagine a packed, crowded room. Suddenly, a group of four people decides to leave together; this is akin to what happens during alpha decay. The original room (nucleus) loses four individuals (nucleons), becoming a different, lighter room (nucleus) with slightly more space.

In mathematical terms, the atomic number of the nucleus goes down by two, and the mass number decreases by four. This implies the element itself changes, moving two places back on the periodic table. For example, uranium-238 decaying into thorium-234. This transition to a more stable state, however, isn't a simple energy shift within the same nucleus, but rather a full-blown identity change—a key point to remember.
Gamma Decay
Gamma decay is a bit different from other types of nuclear decay. Think of it as an excited person calming down without leaving the room. There's no mass migration here; instead, it's all about an internal transition.

In gamma decay, the nucleus goes from a high-energy 'excited' state to a lower-energy 'ground' state, and in doing so, it emits a gamma ray—a form of very high-energy electromagnetic radiation. The nucleus remains the same in terms of element and mass, making this the correct answer to the exercise's question. It's like calming down after jumping around; you're still the same person, just less energetic.

This mode of decay is significant because it doesn't alter the identity of the atom, but it does lower its energy level, which can have implications for the stability and behavior of the material.
Beta Decay
In the world of subatomic shapeshifters, beta decay stands out. During the beta-minus decay process, a neutron is magically transformed into a proton, plus a fast-moving electron and an antineutrino are expelled. Conversely, beta-plus decay sees a proton become a neutron, releasing a positron and a neutrino.

It's like a room changing its function: a library turning into a study room, for example. Although the outside looks the same (the mass remains), the inside is rearranged. With a shift in atomic number, the nucleus essentially rebrands into a new element—one spot over on the periodic table. This type of decay is crucial for elements seeking stability through a more balanced ratio of protons to neutrons.
Nuclear Transitions
Nuclear transitions are the fine adjustments within an atomic nucleus without any drastic changes to its composition or identity. In essence, gamma decay, as discussed earlier, is a prime example of such a transition where there's a shift from a higher to a lower energy state.

Nuclear transitions can be thought of as the various poses a statue might take—it's always the same statue, but the specific pose can vary. It could be a tiny tweak in the structure, or a significant internal reconfiguration, but externally it's still the same figure. These transitions are pivotal for understanding the stability and radioactive properties of an element, and they have wide-reaching implications, from medical imaging techniques to nuclear power generation.

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Most popular questions from this chapter

Why are atomic nuclei more or less limited in size and neutron-proton ratios? That is, why are there no stable nuclei with 10 times as many neutrons as protons, and why are there no atomic nuclei the size of marbles?

Billions of years ago, our Solar System was created out of the remnants of exploding stars. Nuclear scientists believe that two isotopes of uranium, \({ }^{235} \mathrm{U}\) and \({ }^{238} \mathrm{U},\) were created in equal amounts at the time of a stellar explosion. However, today \(99.28 \%\) of uranium is in the form of \({ }^{238} \mathrm{U}\) and only \(0.72 \%\) is in the form of \({ }^{235} \mathrm{U}\). Assuming a simplified model in which all of the matter in the Solar System originated in a single exploding star, estimate the approximate time of this explosion.

Cobalt has a stable isotope, \({ }^{59} \mathrm{Co},\) and 22 radioactive isotopes. The most stable radioactive isotope is \({ }^{60} \mathrm{Co} .\) What is the dominant decay mode of this \({ }^{60}\) Co isotope? a) \(\beta^{+}\) b) \(\beta^{-}\) c) \(\mathrm{P}\) d) \(n\)

In neutron stars, which are roughly \(90 \%\) neutrons and supported almost entirely by nuclear forces, which of the following binding-energy terms becomes relatively dominant compared to ordinary nuclei? a) the Coulomb term b) the asymmetry term c) the pairing term d) all of the above e) none of the above

Before you look it up, make a prediction of the spin (intrinsic spin, i.e., actual angular momentum) of the deuteron, \({ }^{2} \mathrm{H}\). Explain your reasoning.

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