Respuesta :

Core reactions that occur:

¹⁴₇N + ⁴₂He → ¹⁷₈O + ¹₁H

Can be written :

¹⁴₇N (α, H) ¹⁷₈O  , α = ⁴₂He

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.

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

At the core, the 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:

X + a ---> Y + b + Q

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, a fusion reaction occurs with ⁴₂He which produces isotopes of Oxygen and ¹₁H protons.

¹⁴₇N + ⁴₂He → ¹⁷₈O + ¹₁H

This reaction can be written down

¹⁴₇N (α, H) ¹⁷₈O

Learn more

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Keywords: Fusion and fission reactions, alpha ₂He⁴, nuclei

Ver imagen ardni313

The chemical equation 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}}}a[/tex] .

Further Explanation:

An unstable nucleus tends to release energy in order to stabilize itself. This process of energy release is termed as radioactivity. The energy is released in the form of various particles such as alpha, beta and gamma particles.

The helium nucleus consisting of two protons and two neutrons act as an alpha particle. The atomic number of helium is 2 and its mass number is 4. The alpha particle has a charge of +2.

The reactants of the given reaction are 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).

Let us assume A to be the product nuclide. So the given chemical reaction can be written as follows:

[tex]_{\text{7}}^{{\text{14}}}{\text{N}} + _2^4{\text{He}} \to _9^{18}{\text{A}}[/tex]

The product nuclide (A) can also be split to form oxygen-17 and a hydrogen atom. So the above reaction can also be reframed 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:

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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, oxygen-17, hydrogen atom, chemical equation.

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