(a) Compare nuclear density of hydrogen (₁H¹) with its atomic density. (Assume the atom to have the radius of its first Bohr…

(a) For ₁H¹ the nucleus is a single proton. Its nuclear radius is estimated by the nuclear-radius formula R_N = R₀ A^(1/3), with R₀ ≈ 1·2 × 10⁻¹⁵ m and A = 1. Hence R_N ≈ 1·2 × 10⁻¹⁵ m.

Nuclear volume: V_N = (4/3)πR_N³ = (4/3)π(1·2 × 10⁻¹⁵)³ ≈ 7·24 × 10⁻⁴⁵ m³.

Taking the mass as the proton mass, m ≈ 1·67 × 10⁻²⁷ kg, the nuclear density is ρ_N = m/V_N ≈ 1·67 × 10⁻²⁷ / 7·24 × 10⁻⁴⁵ ≈ 2·3 × 10¹⁷ kg m⁻³.

The atomic radius is the first Bohr radius: a₀ ≈ 5·29 × 10⁻¹¹ m.

Atomic volume: V_atom = (4/3)πa₀³ ≈ (4/3)π(5·29 × 10⁻¹¹)³ ≈ 6·20 × 10⁻³¹ m³.

Atomic density: ρ_atom ≈ 1·67 × 10⁻²⁷ / 6·20 × 10⁻³¹ ≈ 2·7 × 10³ kg m⁻³.

Thus ρ_N / ρ_atom = (a₀/R_N)³ ≈ ((5·29 × 10⁻¹¹)/(1·2 × 10⁻¹⁵))³ ≈ 8·6 × 10¹³.

Final comparison: ρ_N ≈ 2·3 × 10¹⁷ kg m⁻³, ρ_atom ≈ 2·7 × 10³ kg m⁻³, so nuclear density is about 10¹⁴ times atomic density. The inference is that the atom is almost entirely empty space; nearly all mass is concentrated in a tiny nucleus, and nuclear matter has an enormous, nearly constant density.

(b) Use Bragg’s law: nλ = 2d sinθ.

Here n = 2, d = 1·41 Å = 1·41 × 10⁻¹⁰ m, and the glancing angle θ = 10°. Therefore λ = 2d sinθ / n = 2d sin10° / 2 = d sin10°.

So λ = 1·41 × sin10° Å = 1·41 × 0·173648 Å = 0·244844 Å.

Thus λ ≈ 0·2448 Å = 2·448 × 10⁻¹¹ m = 24·48 pm.

Final answer: λ ≈ 0·245 Å = 2·45 × 10⁻¹¹ m.

(c) Let R be the deuteron radius, i.e. the distance of each nucleon from the centre of mass. The separation between proton and neutron is then r = 2R.

The reduced mass is μ = m_p m_n/(m_p + m_n) ≈ M/2, where M ≈ 1·67 × 10⁻²⁷ kg, so μ ≈ 8·35 × 10⁻²⁸ kg.

By the uncertainty principle, the uncertainty in relative momentum is Δp ≈ ħ/r = ħ/(2R).

The relative kinetic energy is therefore T ≈ (Δp)²/(2μ) = ħ²/(8μR²).

In the ground state, this zero-point kinetic energy is of the order of the deuteron binding energy B = 2·224 MeV = 2·224 × 1·602 × 10⁻¹³ J ≈ 3·56 × 10⁻¹³ J.

Putting T ≈ B, ħ²/(8μR²) ≈ B, so R ≈ ħ/√(8μB) = ħ/√(4MB).

Now 4MB = 4 × 1·67 × 10⁻²⁷ × 3·56 × 10⁻¹³ ≈ 2·38 × 10⁻³⁹. Hence √(4MB) ≈ 4·88 × 10⁻²⁰ kg m s⁻¹.

Using ħ = 1·055 × 10⁻³⁴ J s, R ≈ 1·055 × 10⁻³⁴ / 4·88 × 10⁻²⁰ ≈ 2·16 × 10⁻¹⁵ m = 2·16 × 10⁻¹³ cm.

Final result: The deuteron radius is of the order of 2·15 × 10⁻¹³ cm, as required. This is the deuteron size or the nucleon’s orbital radius about the centre of mass, not the intrinsic nucleon radius.

(d) The strength of an interaction of elementary particles means the relative magnitude of the coupling constant associated with that interaction. It determines the probability, cross-section, and rate of the corresponding process. A larger coupling constant means a stronger interaction. For comparing forces, one usually expresses the strength as a dimensionless coupling constant at a suitable energy scale.

The four fundamental interactions are classified as follows.

• Strong interaction: This is the strongest interaction. Its coupling constant is of order α_s ≈ 1 at low energy. It acts between quarks through gluons and between hadrons through the residual strong force. Its range is about 10⁻¹⁵ m. It is responsible for nuclear binding.

• Electromagnetic interaction: This is intermediate in strength. Its coupling constant is the fine-structure constant α ≈ 1/137 ≈ 7·3 × 10⁻³. It acts between charged particles and is mediated by photons. Its range is infinite.

• Weak interaction: This is much weaker than the electromagnetic interaction at ordinary energies. Its effective coupling is of order 10⁻⁶–10⁻⁷ in dimensionless terms. It is mediated by W⁺, W⁻, and Z⁰ bosons and has a very short range, about 10⁻¹⁸ m. It is responsible for beta decay and many flavour-changing processes.

• Gravitational interaction: This is the weakest interaction. Between protons, its dimensionless strength is of order 10⁻³⁹. It acts between all masses and energies and is mediated, in quantum theory, by the hypothetical graviton. Its range is infinite.

Thus, in increasing order of strength: gravitational < weak < electromagnetic < strong.

(e) For a type-I superconductor, the critical magnetic field decreases with increasing temperature according to H_c(T) = H_c(0)[1 − (T/T_c)²], valid for 0 ≤ T ≤ T_c. The field is maximum at absolute zero and vanishes at the transition temperature T_c. Similar decreasing behaviour occurs for the lower and upper critical fields of type-II superconductors.

Given: H_c(14 K) = 1·45 × 10⁵ A m⁻¹, H_c(13 K) = 4·2 × 10⁵ A m⁻¹.

Let x = T_c². Then 1·45 × 10⁵ = H_c(0)[1 − 196/x], 4·2 × 10⁵ = H_c(0)[1 − 169/x].

Dividing the first by the second, [1 − 196/x]/[1 − 169/x] = (1·45 × 10⁵)/(4·2 × 10⁵) = 29/84.

Therefore 84(1 − 196/x) = 29(1 − 169/x).

This gives 84 − 16464/x = 29 − 4901/x.

Hence 55 = 11563/x, so x = T_c² = 11563/55 = 210·236 K².

Thus T_c = √210·236 ≈ 14·50 K.

Now H_c(0) = 1·45 × 10⁵ / [1 − 196/210·236].

Since 1 − 196/210·236 ≈ 0·06772, H_c(0) ≈ 1·45 × 10⁵ / 0·06772 ≈ 2·14 × 10⁶ A m⁻¹.

Final answers: T_c ≈ 14·50 K, H_c(0) ≈ 2·14 × 10⁶ A m⁻¹.