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

How much kinetic energy is released and what is the daughter nucleus in the \(\beta^{+}\) decay of nitrogen- \(13 ?\)

Short Answer

Expert verified
The daughter nucleus in the \(\beta^{+}\) decay of nitrogen-13 is \(^{13}_{6}C\), Carbon-13. The kinetic energy release is typically determined experimentally as it requires the precise mass difference, it can also be calculated using E=mc^2 if the mass values are known.

Step by step solution

01

Identify the Parent Nucleus

The parent nucleus is Nitrogen-13. We denote it as \(^{13}_{7}N\), where 13 is the mass number (sum of protons and neutrons) and 7 is the atomic number (number of protons).
02

Identify the Daughter Nucleus in \(\beta^{+}\) Decay

During \(\beta^{+}\) decay, a proton changes into a neutron, a positron (denoted by \(^{0}_{1}\beta^{+}\)), and a neutrino. Therefore, the atomic number decreases by one and the mass number remains unchanged. This results in the formation of Carbon-13 atom, which is denoted by \(^{13}_{6}C\). This is the daughter nucleus.
03

Calculate the Kinetic Energy Released

In this step, using the nuclear reactions we know that a small amount of mass is lost, which gets transformed into energy as per the equation \(E=mc^2\). This lost mass (or mass defect) is the difference between the mass of Nitrogen-13 and the sum of the mass of Carbon-13 and the positron. However, this mass difference is so small that it can be neglected, and the kinetic energy release is often measured directly as a part of the lab experiment. However, if the mass difference were given or could be found from reference values, we could calculate the energy released using \(E = mc^2\) where m is mass and c is the speed of light.

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.

Kinetic Energy Release in Decay
In nuclear reactions, such as beta plus decay, energy is released in the form of kinetic energy. This is because, during the decay process, a small amount of the nucleus's mass is converted into energy. This is guided by the principle of mass-energy equivalence, as described by Einstein's equation, \(E=mc^2\). Here, \(E\) represents energy, \(m\) is the mass lost, and \(c\) is the speed of light.In the case of beta plus decay of Nitrogen-13, when a proton inside the nucleus is transformed into a neutron, the resulting positron and neutrino carry away the kinetic energy. While pinpointing the exact kinetic energy released can be complex without precise mass difference data, it is often measured directly. In practice, this release contributes to the kinetic energy required to power the resulting particles and features observably in lab experiments.
Daughter Nucleus Identification
In beta plus decay, the parent nucleus undergoes transformation where a proton is converted into a neutron, which results in a change in the element's identity. The atomic number, which is the count of protons, decreases by one because of this transformation. However, the mass number, representing the sum of protons and neutrons, remains unchanged.For Nitrogen-13, represented as \(^{13}_{7}N\), during \(\beta^{+}\) decay, one proton is converted into a neutron, resulting in the atomic number moving from 7 to 6. This establishes the daughter nucleus as Carbon-13, denoted as \(^{13}_{6}C\). Understanding the shift in atomic numbers is crucial to identify the daughter nucleus accurately, as it reflects the essential changes happening within the atom during decay.
Mass-Energy Equivalence in Nuclear Reactions
Mass-energy equivalence is a pivotal concept when it comes to nuclear reactions, such as beta plus decay. Einstein's famous equation \(E=mc^2\) illustrates that mass can be converted into energy. During nuclear reactions, the loss of a tiny amount of mass from the system results in a significant amount of energy being released.This concept applies to the beta plus decay of Nitrogen-13, where a small mass loss occurs when a proton changes into a neutron, accompanied by the emission of a positron and a neutrino. The mass lost is fractional yet results in the release of substantial energy that manifests as kinetic energy carried by the decay products. This principle not only helps calculate the potential energy release in theoretical contexts but also guides experimental setups where precise energy measurements are required.

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

An untrained but perceptive friend exclaims, "They say that nuclear energy can be released by sticking nuclei togethcr and by breaking them apart. That doesn't make sense"" Straighten out your friend's confusion.

Potassium. \(40(Z=19 . N=21)\) is a radioactive isotope that is rare but not unknown in nature. It is particularly interesting in that it lies along the curve of stability yet decays by both \(\beta^{+}\) and \(\beta^{-}\) - that is, in both directions away from the curve of stability. (a) Identify the daughter nuclei for both decays. (b) Many factors governing nuclear stability are discussed in the chapter (e.g., those in the semiempirical binding energy formula, magic numbers, and even numbers). Identify those that would argue only for \(\beta^{-}\) decay. (c) Which factors argue only for \(\beta^{+}\) decay? (d) Which argue for either \(\beta^{-}\) or \(\beta^{+}\) decay?

Determine the approximate ratio of the diameter of a uranium nucleus \((A=238)\) to that of a beryllium nucleus \(\\{A=9\rangle\)

In an assembly of fissionable material, the larger the sur. face area per fissioning nucleus (i.e., per unit volume), the more likely is the escape of valuable neutrons. (a) What is the surface-to-volume ratio of a sphere of radius \(r_{0} ?\) (b) What is the surf ace-to-volume ratio of a cube of the same volume? (c) What is the surface-to. volume ratio of a sphere of twice the volume?

Certain nuclei with half-lives between days and a few years are found in nature in small abundances that do not change al all over many, many years. How is this possible? (Hint: Natural uranium and thorium have ren long half- lives.)

See all solutions

Recommended explanations on Physics 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