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Based on a favorable \(N-Z\) ratio for the product nucleus, write the most plausible equation for the decay of \(\frac{14}{6} \mathrm{C}\).

Short Answer

Expert verified
The most plausible equation for the decay of \(\frac{14}{6} \mathrm{C}\) based on a favorable N-Z ratio, which suggests a beta decay, is \(\frac{14}{6} \mathrm{C} \rightarrow \frac{14}{7} \mathrm{N} + -1e + \bar{\nu}\).

Step by step solution

01

Identification of the decay process

The decay of \(\frac{14}{6} \mathrm{C}\) based on a more favorable N-Z ratio indicates beta decay. This is because, in beta decay, a neutron in an atom's nucleus turns into a proton, an electron which is known as beta particle and an electron antineutrino.
02

Writing the decay equation

In the given \(\frac{14}{6} \mathrm{C}\), the upper number 14 is the mass number (A), which is the sum of protons and neutrons in the nucleus, and the lower number 6 is the atomic number (Z) indicating the number of protons. In beta decay, the atomic number increases by 1 (since one neutron becomes an extra proton) and the mass number remains the same (as the total number of protons and neutrons is unchanged). So, the product nucleus after the decay will be \(\frac{14}{7} \mathrm{N}\). Additionally, the emission of the beta particle (\(-1e\)) and an antineutrino (\(\bar{\nu}\)) also need to be represented in the equation.
03

Final equation for beta decay of Carbon-14

Putting all elements together, the most plausible equation for the decay of \(\frac{14}{6} \mathrm{C}\) would be: \(\frac{14}{6} \mathrm{C} \rightarrow \frac{14}{7} \mathrm{N} + -1e + \bar{\nu}\). This equation states that carbon-14 decays to form nitrogen-14, while emitting a beta particle and an antineutrino.

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Key Concepts

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

N-Z Ratio
The N-Z ratio is a crucial concept in nuclear chemistry, especially when discussing stability of atomic nuclei. It represents the number of neutrons (\(N\)) to the number of protons (\(Z\)) in an atomic nucleus. Isotopes with certain N-Z ratios are more stable than others.

For stable isotopes, this ratio stays within a specific range which varies depending on the size of the nucleus. Lighter elements tend to have a ratio close to 1:1, while heavier elements usually have more neutrons than protons, with ratios incrementally increasing. The stability of a nucleus is compromised when the ratio diverges significantly from the stability range.

If a nucleus has too many neutrons, beta decay can occur to transform a neutron into a proton, thereby increasing the Z value while keeping the overall mass number (\(A\)) constant. This process moves the N-Z ratio towards a more stable value, which explains why Carbon-14, with too many neutrons for stability, undergoes beta decay.
Radioactive Decay
Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. This spontaneous transformation can involve the release of alpha particles, beta particles, gamma rays, and conversion electrons, among other particles. In the case of beta decay, which is one type of radioactive decay, a neutron in the nucleus is transformed into a proton and emits a beta particle (an electron \(e^-\)) and an antineutrino (\(\bar{u}\)).

Radioactive decay can be predictable and quantified. The half-life of an isotope - the time it takes for half of all the radioactive nuclei in a sample to decay - is a characteristic property of each radioactive species. Knowing the half-life of Carbon-14, for instance, allows us to perform carbon dating, a technique used to determine the age of archaeological finds.
Carbon-14 Decay
Carbon-14 decay is an example of beta decay, which is utilized in the well-known process of radiocarbon dating. Carbon-14 is a radioactive isotope with a useful property - it is formed in the upper atmosphere and integrates into living organisms. When an organism dies, it stops absorbing Carbon-14, and the isotope starts to decay.

The decay equation for Carbon-14 is: \(\frac{14}{6}\mathrm{C} \rightarrow \frac{14}{7}\mathrm{N} + -1e + \bar{u}\). This indicates that an unstable Carbon-14 nucleus emits a beta particle and an antineutrino as it transforms into the more stable Nitrogen-14 isotope. The C-14 half-life of about 5730 years enables scientists to date organic materials up to about 50,000 years old with a fair degree of accuracy.
Nuclear Chemistry
Nuclear chemistry is the subfield of chemistry dealing with radioactivity, nuclear processes, and nuclear properties. It involves studying various types of radioactive decay, nuclear synthesis, and the effects of ionizing radiation on materials.

Nuclear reactions, unlike chemical reactions, involve changes in an atom's nucleus and often result in the conversion of elements from one type to another. These transformations can release or absorb large amounts of energy, much more than chemical reactions.

Understanding nuclear chemistry is vital for many applications, including medical treatments like radiation therapy, energy generation in nuclear power plants, and assessing the impacts of radiation on biological systems and the environment. The principles of nuclear chemistry also underpin technologies ranging from imaging techniques to the study of the universe through nucleosynthesis in stars.

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Most popular questions from this chapter

5\. In your own words, define the following symbols: (a) \(\alpha ;\) (b) \(\beta^{-} ;\) (c) \(\beta^{+} ;\) (d) \(\gamma ;\) (e) \(t_{1 / 2}\).

Write a plausible equation for the decay of tritium, 3 \(\mathrm{H}\), the radioactive isotope of hydrogen. 1 \(\textrm{ }\).

Of the following nuclides, the one most likely to be radioactive is \((a)^{31} P ;(b)^{66} Z n ;(c)^{35} C l ;(d)^{108} A g\).

For medical uses, radon-222 formed in the radioactive decay of radium-226 is allowed to collect over the radium metal. Then, the gas is withdrawn and sealed into a glass vial. Following this, the radium is allowed to disintegrate for another period, when a new sample of radon- 222 can be withdrawn. The procedure can be continued indefinitely. The process is somewhat complicated by the fact that radon-222 itself undergoes radioactive decay to polonium- 218 , and so on. The half-lives of radium-226 and radon-222 are \(1.60 \times 10^{3}\) years and 3.82 days, respectively.(a) Beginning with pure radium- \(226,\) the number of radon-222 atoms present starts at zero, increases for a time, and then falls off again. Explain this behavior. That is, because the half-life of radon-222 is so much shorter than that of radium- \(226,\) why doesn't the radon-222 simply decay as fast as it is produced, without ever building up to a maximum concentration?(b) Write an expression for the rate of change \((d \mathrm{D} / d t)\) in the number of atoms (D) of the radon- 222 daughter in terms of the number of radium- 226 atoms present initially ( \(\mathrm{P}_{0}\) ) and the decay constants of the parent \(\left(\lambda_{\mathrm{p}}\right)\) and daughter \(\left(\lambda_{\mathrm{d}}\right)\) (c) Integration of the expression obtained in part (b) yields the following expression for the number of atoms of the radon-222 daughter (D) present at a time \(t\).$$\mathrm{D}=\frac{\mathrm{P}_{0} \lambda_{\mathrm{p}}\left(\mathrm{e}^{-\lambda_{\mathrm{p}} \times t}-\mathrm{e}^{-\lambda_{\mathrm{d}} \times t}\right)}{\lambda_{\mathrm{d}}-\lambda_{\mathrm{p}}}$$,Starting with \(1.00 \mathrm{g}\) of pure radium- \(226,\) approximately how long will it take for the amount of radon222 to reach its maximum value: one day, one week, one year, one century, or one millennium?

A sample of radioactive \(\frac{35}{16} \mathrm{S}\) disintegrates at a rate of \(1.00 \times 10^{3}\) atoms \(\min ^{-1} .\) The half-life of \(_{16}^{35} \mathrm{S}\) is \(87.9 \mathrm{d}\) How long will it take for the activity of this sample to decrease to the point of producing (a) \(253 ;\) (b) \(104 ;\) and (c) 52 dis \(\min ^{-1} ?\)

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