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The reaction \({ }_{5} \mathrm{~B}^{8} \longrightarrow \mathrm{Be}^{8}+{ }_{1} \mathrm{e}^{0}\) takes place due to (a) \(\alpha\) decay (b) \(\beta\) decay (c) positron decay (d) electron capture

Short Answer

Expert verified
The reaction is positron decay (option c).

Step by step solution

01

Analyze the Reaction Equation

The given nuclear reaction is \( {}_5 \text{B}^8 \rightarrow {}_4 \text{Be}^8 + {}_{+1} \text{e}^0 \). This shows that Boron-8 is transforming into Beryllium-8 with the emission of an electron with a positive charge.
02

Identify the Decay Process

The emission of an electron with a positive charge in the reaction is indicative of positron emission, commonly known as positron decay. In this process, a proton in the nucleus is converted into a neutron, releasing a positron.
03

Compare Options

Compare the characteristics of different decay types: \( \alpha \)-decay involves emission of helium nuclei, \( \beta \)-decay involves emission of electrons, positron decay involves emission of positrons, and electron capture involves the nucleus absorbing an electron. The given reaction matches positron decay.

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Key Concepts

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

Positron Emission
Positron emission is a fascinating process in nuclear chemistry. It involves the transformation of a proton in the nucleus into a neutron, with the simultaneous emission of a positron. A positron is like an electron, but with a positive charge instead of a negative one. This contrasts with what happens in beta decay, where an electron is emitted. The phenomenon of positron emission is often termed as beta plus (β+) decay due to its similarity with beta decay, but with opposite charge characteristics. This process is crucial in medical imaging, particularly in techniques like Positron Emission Tomography (PET), which takes advantage of the properties of positrons to create images of the body.
Nuclear Decay Processes
Nuclear decay processes involve the transformation of an unstable nucleus into a more stable one. During these processes, the nucleus can emit particles or radiation. These decays include alpha decay, beta decay, gamma decay, and positron emission. Each type of decay process changes the nucleus in different ways:
  • Alpha decay: Emission of helium nuclei (two protons and two neutrons), decreasing the atomic number by 2 and the mass number by 4.
  • Beta decay: Emission of electrons which transforms a neutron into a proton, increasing the atomic number by 1.
  • Positron emission: A proton transforms into a neutron with the emission of a positron, decreasing the atomic number by 1.
  • Electron capture: The nucleus absorbs an electron, converting a proton into a neutron, also decreasing the atomic number by 1.
These processes change the element itself into a different element corresponding to the changes in the nucleus structure.
Nuclear Reactions
Nuclear reactions involve changes in an atom's nucleus and can result in the alteration of one element into another. They are initiated when nuclear components like protons and neutrons are added, emitted, or re-arrange within the nucleus. Different from chemical reactions, nuclear reactions can happen by different modes, such as fission, fusion, and various types of decay, including positron emission.
Nuclear reactions are best known for their critical role in producing energy, both for destructive purposes in nuclear weapons and constructive purposes in nuclear power generation. The energy released in a nuclear reaction is vast compared to chemical reactions, due to changes in the nucleus which are governed by the principle of mass-energy equivalence, famously described by Einstein's equation, \( E=mc^2 \).
Boron-8 Decay
Boron-8 decay is an interesting example of positron emission. In the given reaction, the isotope Boron-8 ( \( {}_5 ext{B}^8 \)) undergoes decay to form Beryllium-8 ( \( {}_4 ext{Be}^8 \)) and a positron ( \( {}_{+1} ext{e}^0 \)). This process illustrates positron emission, where one of the protons in the boron nucleus is transformed into a neutron, emitting a positron as a result.
The transformation from Boron-8 to Beryllium-8 does not change the mass number since the process involves converting a proton to a neutron, and thus the overall mass number of 8 is retained. However, the atomic number decreases from 5 to 4, as there are now fewer protons in the nucleus. This change appropriately reflects the new identity of the nucleus as Beryllium. Such transformations are significant in nuclear science because they demonstrate one of the many ways elements can transmute in nature.

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Most popular questions from this chapter

Half-life period of the radioactive element \(X\) is 10 hours. Amount of \(X\) left in the 11 th hour starting with one \(\mathrm{mol} \mathrm{X}\) is (a) \((1 / 2)^{1 / 10}\) (b) \((1 / 2)^{11 / 10}\) (c) \((1 / 2)^{12 / 11}\) (d) \((1 / 2)^{1 / 11}\)

In a radioactive decay, an emitted electron comes from (a) the nucleus of atom (b) the orbit with principal quantum number 1 (c) the inner orbital of the atom (d) the outermost orbit of the atom

A fresh radioactive mixture containing short lived species \(X\) and \(Y\). Both the species together emitting \(8000 \alpha\) - particles per minute initially. 20 minutes later \(X\) was emitting \(\alpha\) - particles at the rate of 4500 per minute. If the half lives of the species \(\mathrm{X}\) and \(\mathrm{Y}\) are 10 minute and 500 hours, then what is the ratio of initial activities of \(\mathrm{X}\) and \(\mathrm{Y}\) in the mixture?

If a \(\mathrm{X}^{\mathrm{b}}\) species emits firstly a positron, then \(2 \alpha\) and \(2 \beta\) particles and in last \(1 \alpha\) particle is also emitted and finally converts to \(\mathrm{Y}^{\mathrm{d}}\) species, so the correct relation is (a) \(\mathrm{c}=\mathrm{a}-5, \mathrm{~d}=\mathrm{b}-12\) (b) \(\mathrm{c}=\mathrm{a}-5, \mathrm{~d}=\mathrm{b}-10\) (c) \(\mathrm{c}=\mathrm{a}-6, \mathrm{~d}=\mathrm{b}-0\) (d) \(\mathrm{c}=\mathrm{a}-4, \mathrm{~d}=\mathrm{b}-12\)

The half-life of a radioactive isotope is 3 hours. If the initial mass of the isotope was \(300 \mathrm{~g}\), the mass which remained undecayed in 18 hours would be (a) \(4.68 \mathrm{~g}\) (b) \(2.34 \mathrm{~g}\) (c) \(1.17 \mathrm{~g}\) (d) \(9.36 \mathrm{~g}\)

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