Respuesta :
Core reactions that occur:
¹⁴₇N + ⁴₂He → ¹⁷₈O + ¹₁H
Further explanation
Fusion and fission reactions are reactions involving atomic nucleus reactions
Fission reaction is a nucleation division reaction by firing a particle usually neutron so that it gets smaller particles of atomic nuclei.
Example:
Uranium nuclei division reaction
²³⁵₉₂U + ¹₀n -> ⁸⁹₃₆Kr + ¹⁴⁴₅₆Ba + 3¹₀n
In this reaction, Uranium splits into Kr and Ba isotopes and releases energy and gamma radiation
Energy is generated based on the isotope mass from the beginning of the reaction and after division
A fusion reaction is a combining reaction of 2 atomic nuclei. Usually what is often used is a reaction between the hydrogen isotope that will form Helium
³₁Ti + ²₁D -> ⁴₂He + ¹₀n
Isotopes of Hydrogen Tritium and Deuterium form Helium
At the core, reaction applies the law of eternity
- 1. energy
the energy before and after the reaction is the same
- 2. atomic number
the number of atomic numbers before and after the same
- 3. mass number
the number of mass numbers before and after the same
Particles that play a role in core reactions include
- alpha α particles ₂He⁴
- beta β ₋₁e⁰ particles
- gamma particles γ
- positron particles ₁e⁰
In general, the core reaction equation can be written:
[tex]\large{\boxed{\bold{X~+~a~--->~Y~+~b~+~Q}}}[/tex]
X = target core
a = particle fired
Y = new core
b = the particle produced
Q = heat energy
or can be written simply
X (a, b) Y
In the ¹⁴₇N atomic nucleus, it is shot with alpha ⁴₂He rays which produce isotopes of Oxygen and ¹₁H protons.
¹⁴₇N + ⁴₂He → ¹⁷₈O + ¹₁H
This reaction can be written down
¹⁴₇N (α, H) ¹⁷₈O
The sum of the atomic numbers and mass numbers before and after are the same
Learn more
nuclear decay reaction
https://brainly.com/question/4207569
Plutonium − 239
https://brainly.com/question/10125168
exponential decay
https://brainly.com/question/6565665
Keywords: Fusion and fission reactions, alpha ₂He⁴, nuclei
The chemical expression for the given reaction is [tex]\boxed{_{\text{7}}^{{\text{14}}}{\text{N}} + _2^4{\text{He}} \to _8^{17}{\text{O}} + _1^1{\text{H}}}[/tex].
Further Explanation:
Radioactivity is the process of release of energy by an unstable atomic nucleus in the form of different particles such as alpha, beta and gamma particles. It is also known as nuclear or radioactive decay.
An alpha particle is a helium nucleus that consists of two protons and two neutrons. It has a charge of +2.
The given reaction occurs between nitrogen-14 and an alpha particle. Since the alpha particle is a helium nucleus with atomic number 2 and mass number or atomic mass 4, the atomic number of product nuclide becomes 9 (7+2) and its mass number becomes 18 (14+4).
Consider the product nuclide to be X. Therefore the chemical expression for the given reaction is as follows:
[tex]_{\text{7}}^{{\text{14}}}{\text{N}} + _2^4{\text{He}} \to _9^{18}{\text{X}}[/tex] ...... (1)
The product nuclide can also be split to form isotopic oxygen and a hydrogen atom. So equation (1) can also be modified as follows:
[tex]_{\text{7}}^{{\text{14}}}{\text{N}} + _2^4{\text{He}} \to _8^{17}{\text{O}} + _1^1{\text{H}}[/tex]
Learn more:
- What nuclide will be produced in the given reaction? https://brainly.com/question/3433940
- Calculate the nuclear binding energy: https://brainly.com/question/5822604
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Nuclear chemistry
Keywords: alpha particle, atomic number, atomic mass, helium nucleus, nucleus, 2, 4, He, radioactive decay, radioactive disintegration, protons, neutrons, product nuclide, isotopic oxygen, hydrogen atom, chemical expression.