Question

Подсчитайте энергию связи ядра гелия 4/2 HE.

Answer 1

He⁴₂; Np=2- количество протонов в ядре; Nn=4-2=2- количество нейтронов в ядре.
Из справочника:
mp(масса протона)=1.007276 а.е.м(атомная единица массы)
mn(масса нейтрона)=1.008665 а.е.м
Mя(масса ядра)=4.0026 а.е.м.
Eсв=Δm*c²; Δm-дефект массы; c-скорость света в вакууме(3*10⁸ м/с);
Δm=(Np*mp+Nn*mn)-Mя;
Δm=(2*1.007276+2*1.008665)-4.0026=0.029282 а.е.м
1 а.е.м=1.66*10⁻²⁷ кг; Δm=0.029282*1.66*10⁻²⁷=4.860812*10⁻²⁹ кг;
Eсв=4.860812*10⁻²⁹ *(3*10⁸)²=4.37473*10⁻¹² Дж - энергия связи ядра He;
Eсв=4.37473*10⁻¹² Дж/(1.6*10⁻¹⁹ Кл)=27,3*10⁶ эВ=27.3 МэВ
Ответ: Eсв=27.3 МэВ

Answer 2

Calculating the Binding Energy of Helium-4 (4/2 He)

To calculate the binding energy of the helium-4 nucleus (4/2 He), we can use the mass defect and the relationship between mass and energy according to Einstein's mass-energy equivalence principle.

The mass defect (Δm) is the difference between the total mass of the individual nucleons and the mass of the helium-4 nucleus. The binding energy (E) can be calculated using the mass defect and the speed of light (c) according to the formula:

E = Δm * c^2

Let's proceed with the calculation using the given data.

The mass of a proton (p) is approximately 1.0073 atomic mass units (u), and the mass of a neutron (n) is approximately 1.0087 u.

The mass of the helium-4 nucleus (4/2 He) can be calculated as follows: - 2 protons: 2 * 1.0073 u - 2 neutrons: 2 * 1.0087 u

The total mass of the individual nucleons in the helium-4 nucleus is: Total mass = (2 * 1.0073 u) + (2 * 1.0087 u)

Now, we can calculate the mass defect (Δm) and the binding energy (E) using the mass-energy equivalence principle.

Calculation:

Total mass = (2 * 1.0073 u) + (2 * 1.0087 u)

Total mass ≈ 4.0316 u

The actual mass of the helium-4 nucleus is approximately 4.0026 u.

Mass defect (Δm) = 4.0316 u - 4.0026 u

Mass defect ≈ 0.029 u

Using the speed of light (c) as approximately 3.00 x 10^8 m/s, we can calculate the binding energy (E) using the mass defect: E = Δm * c^2

E ≈ 0.029 u * (3.00 x 10^8 m/s)^2

E ≈ 2.61 x 10^-12 Joules

Therefore, the binding energy of the helium-4 nucleus (4/2 He) is approximately 2.61 x 10^-12 Joules.