large surface-to-volume ratio

large surface-to-volume ratio

1. The Scaling Law — Why Surface Dominates at Nano

For a sphere radius $r$:

⇒ $S/V ∝1/r$ — as $r$ shrinks, ratio explodes.

• Bulk $r=1$ mm ⇒ $S/V≈3000$ m⁻¹, surface atoms ~0.0001% (negligible)
• Nano $r=5$ nm ⇒ $S/V≈6×10^8$ m⁻¹, ~40% atoms on surface; at $1$ nm, ~90% surface.

For cube $a$: $S/V=6/a$ same $1/a$ scaling. So shrinking 1000× (µm→nm) increases $S/V$ 1000×.

2. Consequences (why this one property drives many changes)

Consequence	Explanation
High reactivity / catalysis	More dangling bonds, unsatisfied, easy to react (nano Au catalysts, Pt fuel cells)
Lower melting point	Surface atoms have fewer neighbours ⇒ lower binding ⇒ $T_m$ drops as $1/r$ (Gibbs-Thomson: $ΔT_m∝1/r$)
Enhanced adsorption	Sensors, drug delivery rely on large surface for binding
Mechanical: high strength but also surface stress dominates
Sintering at lower $T$	Nano powders sinter far below bulk $T_m$

3. Worked Numbers

Sphere r=10 nm: S/V=3/10e-9=3e8 m⁻¹ ; number atoms N≈(r/a)³ where a≈0.3 nm ⇒ N≈(33)³≈3.6e4 ; surface atoms ≈4πr²·a /V ≈3a/r ≈9% ? Actually ~ (S·a)/V=3a/r≈9% for 10 nm, but with counting more precise ~20% ; at 2 nm, ~60% surface.

4. Exam Points

• State $S/V=3/r$ (sphere) or $6/a$ (cube) and $∝1/r$ — 3-5 marks, very frequent.
• Explain why catalysis and $T_m$ lowering stem from surface-to-volume — 5 marks, very frequent.
• Give numbers: 10 nm ~10–20% surface, 1 nm ~80–90% — 3 marks, frequent.