Question

Identify each of the following as alpha decay, beta decay, positron emission, or gamma emission: a. \({ }_{13}^{27 \mathrm{~m}} \mathrm{Al} \longrightarrow{ }_{13}^{27} \mathrm{Al}+{ }_{0}^{0} \gamma\) b. \({ }_{5}^{8} \mathrm{~B} \longrightarrow{ }_{4}^{8} \mathrm{Be}+{ }_{+1}^{0} e\) c. \({ }_{86}^{220} \mathrm{Rn} \longrightarrow{ }_{84}^{216} \mathrm{Po}+{ }_{2}^{4} \mathrm{He}\)

Short answer

a. Gamma emission; b. Positron emission; c. Alpha decay

Step-by-step solution

Identify the Type of Decay for Part A

Examine the given reaction \({}_{13}^{27 \text{m}} \text{Al} \rightarrow {}_{13}^{27} \text{Al} + {}_{0}^{0} \text{gamma}\)\. In this reaction, the aluminum nucleus transitions from a metastable state \(^{\text{m}}\) to a stable state by emitting a \(^{0}_{0} \text{gamma}\). This type of emission typically involves the release of gamma rays or energy without changing the number of protons or neutrons. Hence, it is identified as gamma emission.

Identify the Type of Decay for Part B

Examine the given reaction \({}_{5}^{8} \text{B} \rightarrow {}_{4}^{8} \text{Be} + {}_{+1}^{0} \text{e}\)\. In this reaction, boron-8 decays to beryllium-8 by emitting a positron \(e^{+}\). This occurs when a proton in the nucleus is converted into a neutron, reducing the atomic number by one. This type of decay is known as positron emission.

Identify the Type of Decay for Part C

Examine the given reaction \({}_{86}^{220} \text{Rn} \rightarrow {}_{84}^{216} \text{Po} + {}_{2}^{4} \text{He}\)\. Here, radon-220 decays to polonium-216 by emitting an alpha particle \(^{4}_{2} \text{He}\), which consists of 2 protons and 2 neutrons. This type of nuclear decay, characterized by the emission of an alpha particle, is identified as alpha decay.

Key concepts

Alpha Decay

In alpha decay, a heavy nucleus loses an alpha particle. An alpha particle, \(^4_2 \text{He}\), is composed of 2 protons and 2 neutrons. During this process, the mass number of the original nucleus decreases by four units, and its atomic number falls by two units. For example, in the reaction \({}_{86}^{220} \text{Rn} \rightarrow {}_{84}^{216} \text{Po} + {}_{2}^{4} \text{He}\), radon-220 transforms into polonium-216 through alpha decay.
As a result, the emitted alpha particle carries away two protons and two neutrons, lowering the atomic mass and the atomic number of the original nucleus.

Beta Decay

Beta decay involves the conversion of a neutron into a proton, or vice versa, within an atomic nucleus. When a neutron converts into a proton, it emits a beta particle (electron) and an antineutrino. This type of decay increases the atomic number by one while the mass number remains unchanged. The equation \({}_{5}^{8} \text{B} \rightarrow {}_{4}^{8} \text{Be} + {}_{+1}^{0} \text{e}\) is an example of beta decay.
Here, boron-8 decays to beryllium-8, releasing a positron (beta particle) \((e^+)\). In this instance, a proton in the nucleus is converted into a neutron. Therefore, beta decay helps nuclei to reach a more stable configuration by altering the ratio of protons to neutrons.

Positron Emission

Positron emission is a type of beta decay where a proton in the nucleus converts into a neutron, emitting a positive electron (positron) and a neutrino. This decreases the atomic number by one while keeping the mass number constant. The given example of \({}_{5}^{8} \text{B} \rightarrow {}_{4}^{8} \text{Be} + {}_{+1}^{0} \text{e}\) represents positron emission.
During this reaction, boron-8 loses a positron, resulting in the formation of beryllium-8. Hence, positron emission helps stabilize the nucleus by adjusting the proton-to-neutron ratio, making the atom more stable.

Gamma Emission

Gamma emission occurs when a nucleus in an excited state releases excess energy in the form of gamma rays. Gamma rays are high-energy photons with no mass and no charge, and they do not alter the number of protons or neutrons in the nucleus. In the reaction \({}_{13}^{27 \text{m}} \text{Al} \rightarrow {}_{13}^{27} \text{Al} + {}_{0}^{0} \text{gamma}\), the metastable aluminum-27 \(^{27 \text{m}} \text{Al}\) converts into a stable aluminum-27 \(^{27} \text{Al}\) by emitting a gamma photon \((^{0}_{0} \text{gamma})\).
The emission of gamma rays allows the nucleus to shift from a higher energy state to a lower one, while maintaining its atomic and mass numbers. This process is common in both alpha and beta decay, where the daughter nucleus often remains in an excited state, needing to release energy to stabilize.