Question

під час бомбардування ядра Літію 7/3Li альфа-частинками вилітає нейтрон. написати рівняння реакції та обчислити її енергетичний вихід​

Answer 1

Ответ:

Объяснение:

Li + He -> B + n (формула з індексами у вкладенні)
M(Li(3;7))=7,016 -3*0,0005486 аом

M(He(2;4))=4,0026 - 2*0,0005486 аом

Mr(B(5;10))=10,01294 -5*0,0005486аом

M(n)=1,00866 аом

ΔM = M(Li)+M(He) - M(B) - M(n) = 7,016+4,0026-10,01294-1,00866

ΔM = -0,003 аом

Δ E= ΔM*931,5 МеВ = -0,003 *931,5 МеВ =-2,79 МеВ

Answer 2

Nuclear Reaction Equation and Calculation of Energy Output during Bombardment of Lithium-7 with Alpha Particles

When lithium-7 (7Li) is bombarded with alpha particles, a nuclear reaction occurs, resulting in the emission of a neutron. The reaction can be represented by the following equation:

7/3Li + α → n + 4/2He

In this equation, the symbol "7/3Li" represents lithium-7, "α" represents an alpha particle (helium-4), "n" represents a neutron, and "4/2He" represents helium-4.

To calculate the energy output of this reaction, we need to consider the mass-energy equivalence principle, as described by Einstein's famous equation, E=mc². This equation states that energy (E) is equal to mass (m) multiplied by the speed of light (c) squared.

To calculate the energy released during the reaction, we need to determine the mass difference between the reactants and the products. The mass of the reactants (7/3Li and α) is greater than the mass of the products (n and 4/2He). This mass difference is converted into energy according to Einstein's equation.

The mass of lithium-7 (7Li) is approximately 7 atomic mass units (amu), and the mass of an alpha particle (α) is approximately 4 amu. The mass of a neutron (n) is approximately 1 amu, and the mass of helium-4 (4/2He) is approximately 4 amu.

To calculate the energy output, we can use the mass difference between the reactants and the products:

Δm = (mass of reactants) - (mass of products) = (7/3Li + α) - (n + 4/2He) = (7 + 4) - (1 + 4) = 6 amu

Using Einstein's equation, we can calculate the energy released:

E = Δm * c²

The speed of light (c) is approximately 3 x 10^8 meters per second.

E = 6 amu * (3 x 10^8 m/s)² = 6 amu * (9 x 10^16 m²/s²) = 54 x 10^16 amu * m²/s²

To convert the energy from atomic mass units (amu) to joules (J), we can use the conversion factor:

1 amu = 1.66 x 10^-27 kg

Since 1 Joule (J) is equal to 1 kilogram (kg) multiplied by (m/s)², we can convert the energy to joules:

E = 54 x 10^16 amu * m²/s² * (1.66 x 10^-27 kg / 1 amu) * (1 J / 1 kg * m²/s²) = 54 x 10^16 * 1.66 x 10^-27 J = 89.64 x 10^-11 J ≈ 8.96 x 10^-10 J

Therefore, the energy output of the reaction is approximately 8.96 x 10^-10 Joules.

Please note that the values used in this calculation are approximate and may vary slightly in different sources.