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

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

Short Answer

Expert verified
The decay equation for tritium would be: \(^3_1H \rightarrow ^3_2He + e^- + \bar{v}_e\), where \(^3_1H\) is tritium, \(^3_2He\) is helium-3, \(e^-\) is the emitted electron (beta particle), and \(\bar{v}_e\) is the electron antineutrino.

Step by step solution

01

Understanding Tritium and Beta Decay

Tritium, 3H, is a radioactive isotope of hydrogen. It undergoes beta decay, in which a neutron turns into a proton and emits an electron (beta particle) and an electron antineutrino.
02

Formulating the Decay Equation

In the case of tritium, one of its neutrons decays into a proton. The atomic number increases by one and the mass number remains the same, creating a new element, helium. The decay also emits an electron, which carries away one unit of negative charge, and an electron antineutrino. The equation for this decay process is:\[ ^3_1H \rightarrow ^3_2He + e^- + \bar{v}_e\]where \(^3_2He\) is helium-3, \(e^-\) is the emitted electron (beta particle) and \(\bar{v}_e\) is the electron antineutrino.

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 type of radioactive decay in which an unstable atomic nucleus transforms into a more stable one by emitting a beta particle. During this process, one of the neutrons in the nucleus is transformed into a proton. As a result, the atom changes from one element to another, moving one place up in the periodic table.

Let's delve deeper into beta decay by looking at an example: tritium decay. In tritium, a neutron decays into a proton, an electron (also referred to as a beta particle), and an antineutrino. This conversion can be represented by the equation:\[^3_1H \to ^3_2He + e^- + \bar{v}_e\]What's noteworthy is that the mass number (top number in the equation) remains constant during beta decay, while the atomic number (bottom number) increases by one. This signifies that the elements' identity changes due to the alteration in the number of protons.
Radioactive Isotopes
Radioactive isotopes, or radioisotopes, are variants of elements that have unstable nuclei. These isotopes decay over time, releasing radiation in the form of particles or electromagnetic waves. Tritium (\(^3_1H\)) is one such example—it's simply hydrogen with two neutrons instead of the usual none.

These isotopes are found in nature but can also be created artificially in nuclear reactors or particle accelerators. The decay of these isotopes is a random process and is quantified by an entity called the half-life, which is the time it takes for half of the isotopes in a sample to decay. Tritium, with a half-life of approximately 12.32 years, is frequently used in scientific research, including studies of the environment and in nuclear fusion experiments.
Neutron Decay
Neutron decay is a process by which a neutron in an atomic nucleus transforms into a proton. This is precisely what happens during the decay of tritium (\(^3_1H\)), a radioactive isotope of hydrogen. This transformation is a fundamental aspect of beta decay, and it is characterized by the emission of two additional particles: an electron and an electron antineutrino.

In this context, it's important to stress that this transformation contributes to the stability of the nucleus. By converting a neutron to a proton, neutrons that are surplus to the stability requirements of the atom are eliminated. Thus, the process of neutron decay is key to understanding how certain isotopes naturally evolve over time.
Electron Antineutrino
The electron antineutrino (\(\bar{v}_e\)) is the almost massless and chargeless 'twin' of the neutrino, which is associated with the electron in particle physics. This particle is emitted along with the electron during beta decay when a neutron turns into a proton. It carries away some of the energy from the decay, ensuring that the laws of conservation of energy, momentum, and angular momentum are obeyed.

Although antineutrinos are incredibly elusive and interact with matter very weakly, they are critically vital in the field of particle physics and nuclear monitoring. They can traverse through planets without being stopped and are so hard to detect that scientists must use incredibly sophisticated and sensitive instruments to observe them.

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

Calculate the minimum kinetic energy (in megaelectronvolts) that \(\alpha\) particles must possess to produce the nuclear reaction $$_{2}^{4} \mathrm{He}+^{14}_{7} \mathrm{N} \longrightarrow^{17}_{8} \mathrm{O}+_{1}^{1} \mathrm{H}.$$ The nuclidic masses are \(_{2}^{4} \mathrm{He}=4.00260 \mathrm{u}\); \(_{7}^{14} \mathrm{He}=14.00307\mathrm{u}\);\(_{1}^{1} \mathrm{H}=1.00783 \mathrm{u}\);\(_{8}^{17} \mathrm{H}=16.99913 \mathrm{u}\);

One method of dating rocks is based on their \(^{87} \mathrm{Sr} /^{87} \mathrm{Rb}\) ratio. \(^{87} \mathrm{Rb}\) is a \(\beta^{-}\) emitter with a half- life of \(5 \times 10^{11}\) years. A certain rock has a mass ratio \(^{87} \mathrm{Sr} /^{87} \mathrm{Rb}\) of \(0.004 / 1.00 .\) What is the age of the rock?

A small quantity of \(\mathrm{NaCl}\) containing radioactive \(_{11}^{24} \mathrm{Na}\) is added to an aqueous solution of \(\mathrm{NaNO}_{3}\). The solution is cooled, and \(\mathrm{NaNO}_{3}\) is crystallized from the solution. Would you expect the \(\mathrm{NaNO}_{3}(\mathrm{s})\) to be radioactive? Explain.

The overall change in the radioactive decay of \({238}_{92} \mathrm{U}\)to 206 \(\mathrm{Pb}\) is the emission of eight \(\alpha\) particles. Show that if \(_{82}^{206} \mathrm{Pb}\)this loss of eight \(\alpha\) particles were not also accompanied by six \(\beta^{-}\) emissions, the product nucleus would still be radioactive.

An ester forms from a carboxylic acid and an alcohol. $$\mathrm{RCO}_{2} \mathrm{H}+\mathrm{HOR}^{\prime} \longrightarrow \mathrm{RCO}_{2} \mathrm{R}^{\prime}+\mathrm{H}_{2} \mathrm{O}.$$ This reaction is superficially similar to the reaction of an acid with a base such as sodium hydroxide. The mechanism of the reaction can be followed by using the tracer \(^{18} \mathrm{O}\). This isotope is not radioactive, but other physical measurements can be used to detect its presence. When the esterifcation reaction is carried out with the alcohol containing oxygen-18 atoms, no oxygen-18 beyond its naturally occurring abundance is found in the water produced. How does this result affect the perception that this reaction is like an acid-base 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