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

Short Answer

Expert verified
Answer: Gamma decay.

Step by step solution

01

Alpha Decay

In alpha decay, the unstable nucleus emits an alpha particle (which is a helium nucleus consisting of 2 protons and 2 neutrons). As a result of this process, the original nucleus loses 2 protons and 2 neutrons, causing a change in both its atomic number and mass number. Hence, alpha decay doesn't involve a transition between states of the same nucleus.
02

Beta Decay

In beta decay, a neutron in the nucleus is converted into a proton or vice versa by emitting either an electron (beta-minus decay) or a positron (beta-plus decay). This results in a change in the atomic number of the nucleus, as the number of protons either increases or decreases by one. Thus, the nucleus changes its identity, and beta decay doesn't involve a transition between states of the same nucleus.
03

Gamma Decay

In gamma decay, the nucleus transitions from an excited state to a lower energy state, releasing energy in the form of a high-energy photon, known as a gamma ray. During this process, there are no changes to the number of protons or neutrons in the nucleus, so the nucleus doesn't change its identity. Thus, gamma decay involves a transition between states of the same nucleus.
04

Answer

The correct answer is c) gamma decay, as it involves a transition between states of the same nucleus without changing its identity.

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Key Concepts

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

Gamma Decay
In understanding nuclear decay modes, gamma decay holds a unique position as it pertains to the energy levels within a nucleus. Unlike alpha and beta decay, gamma decay involves no change in the atomic or mass numbers of an atom. Instead, it occurs when a nucleus in an excited state releases energy to transition back to its ground, or less-energized, state.

This transition is accomplished through the emission of a gamma ray, a form of electromagnetic radiation. Gamma rays are photons with very high energy and consequently, they have a short wavelength on the electromagnetic spectrum. In essence, if we envision an excited nucleus as a tense spring, the emission of a gamma photon is akin to the spring relaxing and liberating energy in the process.

Because there is no change in the number of protons or neutrons, the identity of the element remains the same. An analogy for gamma decay might be a person calming down after an exuberant or agitated state without changing who they are fundamentally.
Alpha Decay
Moving to alpha decay, we delve into a process that results in the reduction of both the mass number and the atomic number of an atom. An alpha particle, which consists of 2 protons and 2 neutrons (the same as a helium-4 nucleus), is expelled from the nucleus of an unstable atom.

This decay mode leads to the transformation of the original element into a new one with an atomic number decreased by two and a mass number decreased by four. For example, when uranium-238 undergoes alpha decay, it becomes thorium-234. The equation for alpha decay is given by:\[ _{Z}^{A}X \rightarrow _{Z-2}^{A-4}Y + _{2}^{4}He \]

The loss of an alpha particle is a significant event for the nucleus as it results in a marked shift in its composition and therefore leads to the transmutation of the element altogether.
Beta Decay
On the topic of transformative decay modes, beta decay alters the composition of the nucleus by changing a neutron into a proton or vice versa. This is achieved through the emission of a beta particle, which can be either an electron (in beta-minus decay) or a positron (in beta-plus decay).

In beta-minus decay, the equation is typically represented as:\[ _{Z}^{A}X \rightarrow _{Z+1}^{A}Y + e^- + \overline{u}_e \]
whereas in beta-plus decay, it is:\[ _{Z}^{A}X \rightarrow _{Z-1}^{A}Y + e^+ + u_e \]

The result of beta decay is a new element with an atomic number shifted by one, reflecting the conversion of a neutron into a proton or a proton into a neutron. This fundamental change in the nucleus means that beta decay does not just involve a transition between states of the same nucleus but rather a transformation into a different nucleus altogether.
Nuclear Transitions
Nuclear transitions cover a broad range of processes that can occur in the nucleus of an atom. These transitions can result in the atom moving from a higher to a lower energy state or from one nuclear configuration to another. The types of nuclear transitions include gamma decay, as discussed, but also other processes like nuclear fission, fusion, and neutron capture.

Each of these transitions is characterized by a change in the nucleus' energy or composition. For instance, nuclear fission involves the splitting of a heavy nucleus into lighter nuclei along with the release of a significant amount of energy. In contrast, nuclear fusion is the combining of light nuclei to form a heavier nucleus, which powers the sun and stars.

It is essential to differentiate between the types of nuclear transitions since some, like gamma decay, merely involve an energy release without a change in nuclear identity, while others can fundamentally alter the composition and identity of the initial nucleus.

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Most popular questions from this chapter

Show that for the case of nuclei with odd mass number, \(A,\) the Bethe- Weizsäcker formula can be written as a quadratic in \(Z\) -and thus, for any given \(A\), the binding energies of the isotopes having that \(A\) take a quadratic form, \(B=a+b Z+c Z^{2}\). Use your result to find the most strongly bound isotope (the most stable one) having \(A=117\).

Assuming that carbon makes up \(14 \%\) of the mass of a human body, calculate the activity of a 75 -kg person considering only the beta decays of carbon- 14

Write equations for the \(\beta^{-}\) decay of the following isotopes: a) \(\frac{60}{27} \mathrm{Co}\) b) \({ }_{1}^{3} \mathrm{H}\) c) \({ }_{6}^{14} \mathrm{C}\)

Physicists blow stuff up better than anyone else. The measure for gauging the usefulness of blowing something up is the fraction of initial rest mass converted into energy in the process. Looking up the necessary data, calculate this fraction for the following processes: a) chemical combustion of hydrogen: \(2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) b) nuclear fission: \(n+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{36}^{89} \mathrm{Kr}+{ }_{56}^{142} \mathrm{Ba}+5 n\) c) thermonuclear fusion: \({ }_{3}^{6} \mathrm{Li}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{4}^{7} \mathrm{Be}+n\) d) decay of free neutron: \(n \rightarrow p+e^{-}+\bar{\nu}_{e}\) e) decay of muon: \(\mu^{-} \rightarrow e^{-}+\nu_{\mu}+\bar{\nu}_{e}\) f) electron-positron annihilation: \(e^{-}+e^{+} \rightarrow 2 \gamma\)

A 42.58 -MHz photon is needed to produce nuclear magnetic resonance in free protons in a magnetic field of \(1.000 \mathrm{~T}\). What is the wavelength of the photon, its energy, and the region of the spectrum in which it lies? Could it be harmful to the human body?

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