Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The nucleus of nitrogen- 18 lies above the stability belt. Write an equation for a nuclear reaction by which nitrogen- 18 can achieve stability.

Short Answer

Expert verified
The nuclear reaction for Nitrogen-18 achieving stability is: \(^{18}_{7}\)N --> \(^{18}_{8}\)O + β- + \(\bar{ν}\), which indicates the beta decay process.

Step by step solution

01

Understanding instability of Nitrogen-18

Nitrogen-18 is above the belt of stability. This means it has too many neutrons compared to protons, making it unstable. To gain stability, it needs to convert some of its excess neutrons into protons through a process known as beta decay.
02

Beta Decay

In beta decay, a neutron in the nucleus of the atom changes into a proton. It emits a beta particle, which is a high-energy electron, and an electron antineutrino.
03

Writing the Nuclear Reaction

In the case of Nitrogen-18 (\(^{18}_{7}\)N), one neutron will change into a proton by emitting a beta particle (β-). Therefore, the Nitrogen-18 transforms into an atom of Oxygen-18 (\(^{18}_{8}\)O). The nuclear reaction can be written as follows: \(^{18}_{7}\)N --> \(^{18}_{8}\)O + β- + \(\bar{ν}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Beta Decay
Beta decay is a fascinating nuclear process that helps an unstable nucleus achieve stability. This occurs when a neutron within the nucleus converts into a proton. During this transformation, a beta particle, which is essentially a high-energy electron, is emitted from the nucleus. Along with the beta particle, a very light and neutral particle called an electron antineutrino is also emitted.
Beta decay helps adjust the number of protons and neutrons, allowing the atom to be more balanced and closer to the atomic stability line.
This process is crucial in many nuclear reactions and is often involved in radioactive decay in various isotopes.
Nitrogen-18
Nitrogen-18 is an isotope of nitrogen that finds itself above the stability belt, meaning it is prone to radioactive decay.
In the realm of nuclear chemistry, this means that nitrogen-18 possesses a higher number of neutrons compared to protons than what is considered stable.
The excess of neutrons makes nitrogen-18 unstable and forces it to undergo beta decay to shed this extra weight.
With beta decay, nitrogen-18 will lose a neutron, which becomes a proton, converting the atom into oxygen-18.
  • Original Element: Nitrogen-18 (\(^{18}_{7}\)N)
  • Transforming into: Oxygen-18 (\(^{18}_{8}\)O)
This conversion helps stabilize the atom, effectively balancing its nuclear forces and placing it back in line with the stability belt.
Nuclear Reactions
Nuclear reactions are not just limited to powering ancient stars and massive reactors; they play a significant role in smaller, more precise atomic changes. These transformations involve the nucleus, altering the composition from one element to another.
A classic example is the transformation observed in nitrogen-18's beta decay, where its nuclear configuration changes, resulting in a different element altogether—oxygen-18.
The reaction can be depicted as:
  • Start: \(^{18}_{7}\)N
  • Process: Emission of a beta particle and an electron antineutrino
  • End: \(^{18}_{8}\)O
This specific type of nuclear reaction showcases how unstable isotopes adjust themselves, achieving stability through the conversion of neutrons to protons, which highlights the incredible dynamics within atomic structures.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What makes water particularly suitable for use as moderator in a nuclear reactor?

In 1997 a scientist at a nuclear research center in Russia placed a thin shell of copper on a sphere of highly enriched uranium-235. Suddenly, there was a huge burst of radiation, which turned the air blue. Three days later, the scientist died of radiation damage. Explain what caused the accident. (Hint: Copper is an effective metal for reflecting neutrons.)

Consider this redox reaction: $$ \begin{array}{r} \mathrm{IO}_{4}^{-}(a q)+2 \mathrm{I}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \\ \mathrm{I}_{2}(s)+\mathrm{IO}_{3}^{-}(a q)+2 \mathrm{OH}^{-}(a q) \end{array} $$ When \(\mathrm{KIO}_{4}\) is added to a solution containing iodide ions labeled with radioactive iodine- \(128,\) all the radioactivity appears in \(\mathrm{I}_{2}\) and none in the \(\mathrm{IO}_{3}^{-}\) ion. What can you deduce about the mechanism for the redox process?

As a result of being exposed to the radiation released during the Chernobyl nuclear accident, the dose of iodine- 131 in a person's body is \(7.4 \mathrm{mC}(1 \mathrm{mC}=1 \times\) \(10^{-3} \mathrm{Ci}\) ). Use the relationship rate \(=\lambda N\) to calculate the number of atoms of iodine- 131 to which this radioactivity corresponds. (The half-life of \({ }^{131} \mathrm{I}\) is 8.1 days.)

A radioactive substance undergoes decay as: $$ \begin{array}{cc} \text { Time (days) } & \text { Mass (g) } \\ \hline 0 & 500 \\ 1 & 389 \\ 2 & 303 \\ 3 & 236 \\ 4 & 184 \\ 5 & 143 \\ 6 & 112 \end{array} $$ Calculate the first-order decay constant and the halflife of the reaction.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free