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7n14 is a strange spelling.
But you think probable to nitrogen.
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∙ 11y ago
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WHAT IS 7N14?
its sniper ammo
What was the reactant that underwent beta decay if 14 7 N was the product?
7N14 is the product of beta- decay of 6C14. Remember, beta- decay involves changing a neutron into a proton, with the emission of an electron and an electron antineutrino. The W- boson is an intermediate product of the changing of a down quark to an up quark that is represented by the change of a neutron into a proton, which then decays into the electron and electron antineutrino pair.. In a nutshell, with beta- decay, atomic number goes up by one, and atomic mass number stays the same.
Examples of isobars in chemistry?
Isobars 7N14, 6C14 2He3, 1H3 92U239, 93Np239
How do you find the isotope in alpha and beta decay?
In Beta- decay, a neutron is converted into a proton, and an electron and electron anti-neutrino are emitted. The Atomic Number goes up by one, and the Atomic Mass Number stays the same. For instance, 6C14 becomes 7N14 plus one electron and one electron anti-neutrino. In Beta+ decay, a proton is converted into a neutron, and a positron and electron neutrino is emitted. The Atomic Number goes down by one, and the Atomic Mass Number stays the same. For instance, 6C11 becomes 5B11 plus one positron and one electron neutrino. Isotopes that decay by Beta+ decay also tend to decay by Electron Capture, a process where an inner K shell electron is absorbed by the nucleus, changing a proton into a neutron and emitting a neutrino. The isotope conversion process would be the same as for Beta+, above. In Alpha decay, a Helium nucleus (two protons and two neutrons) are emitted. The Atomic Number goes down by two, and the Atomic Mass Number goes down by four. For instance, 92U238 becomes 90Th234 plus one Helium nucleus
What happens when carbon 14 undergoes radioactive decay?
6C14 ---------> 7N14 + -1 e0 Beta particle is emitted and carbon changes into nitrogen
How do write equation for alpha decay of actinium-225?
An alpha particle consists of 4 nucleons, which are 2 protons and 2 neutrons. In alpha decay, an alpha particle is emitted from the nucleus of an atom, so the atom loses 2 protons, and a total of 4 nucleons. The atomic number of an atom undergoing alpha decay is reduced by 2, the number of protons lost, and the mass number is reduced by 4, the number of nucleons lost.
What type of particle is emitted when carbon-14 decays into nitrogen-14?
A beta particle. This is because the mass number is the same, so it cannot involve a neutron or a proton. Beta particles can either be a positron or an electron. A positron is just a positive electron. Because the atomic masses are the same, but the atomic number is different, you need an electron or positron, to negate the difference.
What is the equation for beta decay of carbon-14?
14C --> 14N + e-
Does carbon decay by alpha particle emission?
No. Carbon-14 decays by Beta- decay, which ultimately emits an electron, otherwise called a beta particle. Technically, 6C14 decays to 7N14 by the conversion of a neutron into a proton, by the emission of a W- boson, which then (almost immediately) decays into an electron and an electron antineutrino. To understand the conversion of a neutron into a proton, realize that neutrons and protons are made up of quarks, three specifically. The neutron has two down quarks and one up quark, while the proton has one down quark and two up quarks. The W- boson, mediated by the weak atomic force, converts one of the down quarks into an up quark.
What are the steps of radioactive decay through which a beta particle is released?
Beta decay is a type of radioactive decay that comes in two types. The beta plus decay and beta minus decay can be described by use of an equation that places an unstable atom on one side and the products of the beta decay on the other. The beta minus decay of carbon-14 is just a single example of this equation, and here it is: 614C => 714N + e- + ve To learn more about beta decay, use the link below to the related question, "What is beta decay?"
What kind of nuclear reaction would result in the release of a beta particle?
Beta particle( electron having nuclear origin) is emitted when a neutron decays into a proton by giving out electron. The electron produced escapes as a beta particle leaving proton in the nucleus of atom. 0n1 --> 1p1 + -1e0 ( 1e0 is the emitted beta particle) here subscripts denote charge and superscript denote mass in atomic mass unit(amu). Such neutron decay are shown by some radioactive elements. Usually when the n/p (neutron/proton) ratio is higher than required nuclei emit beta particle. Many examples of this type of decay can be given like: 6c14 --> 7N14 + -1e0 (this carbon isotope is used in carbon dating). 90Th232 + 0n1 --> 90Th232 - -1e0 --> 91Pa233 - -1e0 --> 92U233 (this reaction is used in breeder reactors for production of fissile uranium isotope)
What are the stages of thermonuclear fusion that occurs in the Sun's core?
There are two theories for the thermonuclear cycle. In both, the overall reaction is 4 hydrogen nuclei combining to give 1 helium nucleus, 2 positrons and appx. 26.7 MeV of energy.Proton - Proton Cycle:This cycle can be explained in following steps:Step I: When two protons fuse, they form deuteron nucleus (1D2 or 1H1), a positively charged electron called as positron (1e0) and some energy Q1 is released.1H1 + 1H1 ---Fusion--> 1D1 + 1e0 + Q1Step II: The deuteron then fuses with another proton to give 2He3 (i.e. a light isotope of Helium) and γ-quantum radiations.1D2 + 1H1 ---Fusion--> 2He3 + γStep III: For the reaction in Step III to occur, the first two step reactions must occur twice to get two light isotopes of Helium. These two light isotopes fuse to produce stable 2He4.2He3 + 2He3 ---Fusion--> 2He4 + 21H1 + Q2The overall reaction is as follows:41H1 ---Fusion--> 2He4 + 21e0 + 2γ + (2Q1 + Q2)Total energy released = (2Q1+Q2) = 26.7 MeV.Carbon - Nitrogen Cycle:Step I: A proton of Hydrogen first interacts with C12 nucleus to produce N13 isotope and Q1 amount of energy.6C12 + 1H1 ---Fusion--> 7N13 + Q1Step II: Nitrogen N13 decays due to its radioactive nature and emits a positron 1e0.7N13 ---Radioactive decay--> 6C13 + 1e0Step III: Stable C13 nucleus reacts with another proton, releasing N14 and Q2 amount of energy6C13 + 1H1 ---Fusion--> 7N14 + Q2Step IV: Stable N14 now fuses with third proton to form O15 with Q3 amount of energy.7N14 + 1H1 ---Fusion--> 8O15 + Q3Step V: The O15 nucleus decays to give positron.8O15 ---Radioactive decay--> 7N15 + 1e0Step VI: Finally N15 nucleus interacts with forth proton to produce He Nucleus.7N15 + 1H1 ---Fusion--> 2He4 + 6C12The overall reaction is as follows:41H1 ---Fusion--> 2He4 + 21e0 + (Q1 + Q2 + Q3)Total energy released = (Q1+Q2 Q3) = 26.7 MeV.
