(a) What is the minimum energy required to break a ₂He⁴ nucleus into free protons and neutrons? [ Given, m_H = 1·007825 amu, m_n…

(a)

The minimum energy to break ₂He⁴ into free protons and neutrons is its nuclear binding energy. Use the mass-defect method.

Let m_H be the atomic hydrogen mass, which includes one proton plus one electron. The helium atomic mass includes two electrons. Hence electron masses cancel in:

Δm = (2 m_H + 2 m_n) − m_He

Substitute:

Δm = 2(1.007825) + 2(1.008665) − 4.002603 amu = 2.015650 + 2.017330 − 4.002603 amu = 0.030377 amu

Using 1 amu = 931.5 MeV/c²,

E_b = Δm c² = 0.030377 × 931.5 MeV = 28.296 MeV ≈ 28.30 MeV

Minimum energy = 28.30 MeV. This is the threshold energy; any extra energy appears as kinetic energy of the free protons and neutrons.

(b)(i)

For a uniformly charged sphere of charge Ze and radius R, the electrostatic self-energy is

U = (3/5)(Ze)²/(4π ε₀ R)

Use e²/(4π ε₀) = 1.44 MeV fm. Given R = 1.2×10⁻¹³ ∛A cm = 1.2 ∛A fm.

Initial U-236 nucleus: Z_i = 92, A_i = 236.

R_i = 1.2 ∛236 fm ≈ 1.2 × 6.1797 fm = 7.4157 fm

U_i = (3/5)(1.44)(92²)/(7.4157) MeV = 0.6 × 1.44 × 8464/7.4157 MeV = 986.14 MeV

Each equal fragment has Z_f = 46, A_f = 118.

R_f = 1.2 ∛118 fm ≈ 1.2 × 4.9049 fm = 5.8858 fm

U_f,each = (3/5)(1.44)(46²)/(5.8858) MeV = 0.6 × 1.44 × 2116/5.8858 MeV = 310.61 MeV

Total final electrostatic energy:

U_f = 2 × 310.61 MeV = 621.23 MeV

Reduction:

ΔU = U_i − U_f = 986.14 − 621.23 MeV = 364.91 MeV ≈ 3.65×10² MeV

Equivalently, since A_f = A_i/2, ΔU = U_i(1 − 2^(−2/3)), which gives the same value. This is the self-energy decrease at fission; the subsequent Coulomb repulsion of the separating fragments appears as kinetic energy.

(b)(ii)

No. A free electron cannot absorb a photon completely because energy and momentum cannot both be conserved in vacuum.

Assume the electron is initially at rest. If the photon is completely absorbed, energy conservation gives

hν + m_e c² = γ m_e c²

and momentum conservation gives

hν/c = γ m_e v

From momentum, the electron momentum is p = hν/c. Its total energy must then be

E_e = √((hν)² + m_e² c⁴)

But energy conservation requires

E_e = hν + m_e c²

Squaring gives

(hν)² + m_e² c⁴ = (hν)² + 2hν m_e c² + m_e² c⁴

which is impossible unless hν = 0. Hence a free electron cannot absorb all the photon energy. In Compton scattering, the photon transfers only part of its energy and is scattered. Full transfer is possible only in the presence of a third body, such as a nucleus, which takes recoil momentum.

(c)

Hysteresis in ferromagnetic materials arises from their domain structure. A ferromagnetic specimen contains many magnetic domains, each spontaneously magnetized in some easy direction. When an external magnetic field H is applied, domains favourably oriented with H grow, while others shrink; domain walls move. The magnetization M, and hence B, therefore increases.

However, domain-wall motion is not perfectly reversible. Impurities, grain boundaries, dislocations, vacancies, residual stresses, and magnetocrystalline anisotropy pin the domain walls. Extra energy is needed to overcome this pinning, so wall motion becomes irreversible. When H is reduced, the domains do not return to their original configuration. Some magnetization remains even at H = 0; this is remanence. To reduce B to zero, a reverse field called the coercive force must be applied. Thus B lags behind H, producing the hysteresis loop.

The area enclosed by the hysteresis loop represents the energy dissipated as heat per unit volume per magnetization cycle. It is given by

W = ∮ H dB

per unit volume. Its unit is J/m³ per cycle. A large area means large hysteresis loss, high coercivity, and a hard magnetic material, useful for permanent magnets. A small area means low hysteresis loss and a soft magnetic material, suitable for transformer cores and electromagnets. Thus the loop area is a direct measure of irreversible magnetic energy loss.