(a) Establish the Rutherford's scattering cross section formula for α-particle by considering the standard assumptions and symbols. 20 marks

(b) By assuming the nucleus as a cubical box of length equal to the nuclear diameter 10⁻¹² cm, calculate the

Question

(a) Establish the Rutherford's scattering cross section formula for α-particle by considering the standard assumptions and symbols. 20 marks

(b) By assuming the nucleus as a cubical box of length equal to the nuclear diameter 10⁻¹² cm, calculate the kinetic energy of the highest level occupied nucleon of iron-56 nucleus. 15 marks

(c) What do you understand by nuclear forces ? Explain meson theory of exchange forces. 5+10=15 marks

UPSC Answer Check · Accepted Answer

(a) Let the α-particle have charge q₁ = z e and the nucleus q₂ = Z e. Assume a point Coulomb field V(r) = k q₁q₂/r, with k = 1/(4π ε₀), target infinitely heavy, non-relativistic single scattering, impact parameter b, incident kinetic energy E = ½ m v₀². Angular momentum is L = m v₀ b = m r²(dφ/dt).

Use the Binet orbit equation for a central force:
u'' + u = - m F(1/u)/(L² u²), u = 1/r.
For the repulsive Coulomb force F = k q₁q₂/r² = k q₁q₂ u², so
u'' + u = - m k q₁q₂/L².
Put α = m k q₁q₂/L². The solution is
u = α(ε cos φ − 1),
where ε is the eccentricity. At infinity u = 0, so ε cos φ₀ = 1, where φ₀ is the asymptote angle. For repulsive scattering, θ = π − 2φ₀, hence
sin(θ/2) = cos φ₀ = 1/ε.

The energy is
E = (L²/2m)(u'² + u²) + k q₁q₂ u.
Substituting u = α(ε cos φ − 1) and simplifying gives
ε² − 1 = 2 E L²/(m (k q₁q₂)²).
Since E = ½ m v₀² and L = m v₀ b,
ε² − 1 = (m v₀² b/(k q₁q₂))².
Therefore
cot(θ/2) = √(ε² − 1) = m v₀² b/(k q₁q₂),
so
b = k q₁q₂/(m v₀²) cot(θ/2) = q₁q₂/(8π ε₀ E) cot(θ/2).

For central scattering,
dσ/dΩ = (b/sinθ)|db/dθ|.
With A = q₁q₂/(8π ε₀ E), b = A cot(θ/2), db/dθ = −(A/2)csc²(θ/2).
Thus
dσ/dΩ = [A cot(θ/2)/sinθ]·[(A/2)csc²(θ/2)] = A²/(4 sin⁴(θ/2)).
Hence
dσ/dΩ = [q₁q₂/(16π ε₀ E)]² csc⁴(θ/2).
For α-particle q₁ = 2e, q₂ = Z e:
dσ/dΩ = [Z e²/(8π ε₀ E)]² csc⁴(θ/2).
This is Rutherford’s scattering cross-section formula. It is valid for classical, non-relativistic, single Coulomb scattering by a very heavy nucleus.

(b) For a particle in a cubical box of side L,
E(n₁,n₂,n₃) = [π² ħ²/(2mL²)](n₁² + n₂² + n₃²), n₁,n₂,n₃ = 1,2,3,…
Let E₀ = π² ħ²/(2mL²). Given L = 10⁻¹² cm = 10⁻¹⁴ m. Using m ≈ 1.67×10⁻²⁷ kg,
E₀ = [π²(1.055×10⁻³⁴)²]/[2(1.67×10⁻²⁷)(10⁻¹⁴)²] J
≈ 3.28×10⁻¹³ J = 2.05 MeV.

For iron-56, Z = 26 protons and N = 30 neutrons. Protons and neutrons are distinct; each spatial level accepts two spin orientations. Order levels by S = n₁² + n₂² + n₃²:
S = 3: 1 state, capacity 2, cumulative 2
S = 6: 3 states, capacity 6, cumulative 8
S = 9: 3 states, capacity 6, cumulative 14
S = 11: 3 states, capacity 6, cumulative 20
S = 12: 1 state, capacity 2, cumulative 22
S = 14: 6 states, capacity 12.
For 26 protons, 4 states of S = 14 are filled; for 30 neutrons, 8 states of S = 14 are filled. Hence the highest occupied level has S = 14.
The highest energy is E = 14E₀ = 14 × 2.05 MeV ≈ 28.7 MeV.
This assumes independent nucleons in an infinite square well and neglects residual nuclear interaction.

(c)(i) Nuclear forces are the strong residual interactions between nucleons that bind protons and neutrons inside the nucleus. Their main features are: short range of order 10⁻¹⁵ m; strong attraction at intermediate distances; a very short-range repulsive core; charge independence and charge symmetry; spin dependence and tensor character; saturation, so nuclear density and binding energy per nucleon remain nearly constant; and exchange character, since they arise from exchange of mesons.

(c)(ii) Yukawa’s meson theory explains nuclear forces as exchange forces. A nucleon emits a virtual meson and another absorbs it, producing an exchange interaction. The range R is fixed by the uncertainty principle:
R ≈ ħ/(m c),
where m is the exchanged meson mass. For R ≈ 1.4 fm, m c² ≈ 140 MeV, corresponding to the pion. Thus charged pions π⁺ and π⁻ give charge-exchange reactions such as
n → p + π⁻, p + π⁻ → n,
while neutral pion exchange gives p–p or n–n interactions. The resulting Yukawa potential is
V(r) = −(g²/4π) exp(−μr)/r, μ = mπ c/ħ,
which gives the correct long-range tail of the nuclear force. Because pions are pseudoscalar particles, their coupling involves spin and isospin operators, yielding spin-dependent and tensor components essential for the deuteron quadrupole moment. Heavy mesons such as ρ and ω exchange account for shorter-range attraction and the strong repulsive core, ensuring nuclear saturation. In this way the exchange of mesons explains both the attractive long-range part and the short-range repulsive behavior of nuclear forces.

(a) Establish the Rutherford's scattering cross section formula for α-particle by considering the standard assumptions and…

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