(a) If the nuclear force is charge independent and a neutron and proton form a bound state then why is there no bound state for…

(a) Why there is no bound dineutron

Charge independence means that the nuclear force between two nucleons in the same quantum state (same spin, orbital and isospin state) is the same for pp, nn and np.

• The deuteron (np) is bound only in the spin-triplet state (S = 1, mainly L = 0 with about 4% L = 2), with a binding energy of 2.22 MeV. The np spin-singlet state (S = 0) is not bound (only a virtual state).
• Two neutrons are identical fermions, so the total wavefunction must be antisymmetric. In the lowest spatial state (L = 0, symmetric) the spins must be antisymmetric, so only the singlet S = 0 is allowed. The triplet S = 1 would need L odd (L = 1), where the centrifugal barrier and weaker force prevent binding.
• Since the np singlet is itself unbound, charge independence predicts that the nn singlet (and pp singlet) are also unbound, which is what is observed. In isospin language, the deuteron is the T = 0 state, while nn, pp and the np singlet form the T = 1 triplet, none bound.

Information on the nucleon-nucleon force:

1. It is spin dependent: stronger in the triplet than in the singlet state.
2. It has a non-central tensor component, shown by the deuteron’s quadrupole moment.
3. It is only just strong enough to bind the deuteron and just too weak to bind the singlet, so it is short-ranged with a depth of a few tens of MeV.

(b) Quark-model prohibitions

(i) A baryon of spin 1: a baryon is made of three quarks (qqq), each of spin ½. Two spins give S = 0 or 1, adding the third gives S = ½ or 3/2. A system of three fermions always has half-integer spin, so a baryon is a fermion and spin 1 is impossible.

(ii) An anti-baryon of charge +2: an anti-baryon is three antiquarks. The antiquark charges are −⅔ (ū, c̄, t̄) and +⅓ (d̄, s̄, b̄). The largest possible charge is d̄d̄d̄ (or any three of d̄, s̄, b̄) = +1, as in the anti-Δ⁻. A charge of +2 would need three antiquarks of charge +⅔ each, which do not exist. The allowed anti-baryon charges are −2, −1, 0 and +1 (mirror images of the baryon charges +2, +1, 0, −1, as for Δ⁺⁺).

(c) Type-II versus Type-I superconductors for magnets

• Type-I (Pb, Hg, Sn): show complete Meissner effect up to a single critical field H_c, which is small (below about 0.1 T). Above it superconductivity is lost. Currents flow only in a thin surface layer, so they cannot carry large currents in strong fields.
• Type-II (NbTi, Nb₃Sn, YBCO): perfect diamagnetism up to H_c1, then a mixed (vortex) state up to a very high H_c2 (about 10 T for NbTi, about 25-30 T for Nb₃Sn, above 100 T for YBCO). Flux enters as quantised Abrikosov vortices (Φ₀ = h/2e) while the rest stays superconducting.
• Flux pinning at defects and impurities prevents vortex motion under the Lorentz force, so there is no dissipation and the wires carry high critical current density in high fields.
• Hence magnets for MRI, NMR, particle accelerators (LHC uses NbTi at 8.3 T), fusion devices and maglev all use Type-II materials.

(d) FET as a unipolar transistor and its advantages

• Unipolar: in an FET the current is carried by majority carriers of one type only, electrons in an n-channel or holes in a p-channel. The gate voltage controls the channel width (JFET) or induces the channel (MOSFET). In a BJT both electrons and holes take part (minority carriers diffuse across the base), so it is bipolar.
• Advantages over BJT:

1. Very high input impedance (about 10⁸-10¹⁴ Ω, gate current negligible), voltage-controlled, so little loading of the source.
2. Low noise, since there are no junction recombination effects in the channel.
3. Better thermal stability: channel resistance increases with temperature (negative temperature coefficient of current), so there is no thermal runaway.
4. No minority-carrier storage, no offset voltage when used as a switch.
5. Smaller size and simpler fabrication, so MOSFETs dominate integrated circuits and low-power CMOS logic.

(e) Universal gates and the X-OR gate

NAND and NOR are universal because any Boolean function can be built using only one of them:

• NOT: NAND (or NOR) with inputs tied, Y = (A·A)′ = A′.
• AND: NAND followed by a NAND-inverter, ((AB)′)′ = AB.
• OR: invert A and B with NANDs, then NAND them, (A′B′)′ = A + B (De Morgan).
• Similarly NOR gives NOT, OR, and AND by duality.

X-OR gate: output 1 when the inputs differ.

• Boolean equation: Y = A ⊕ B = A′B + AB′ = (A + B)(A′ + B′).
• Truth table (A, B → Y): 0, 0 → 0; 0, 1 → 1; 1, 0 → 1; 1, 1 → 0.
• Using four NAND gates: G1 = (AB)′, G2 = (A·G1)′ = A′ + B, G3 = (B·G1)′ = A + B′, Y = (G2·G3)′ = A ⊕ B.

X-OR is used in half and full adders, comparators and parity generators.