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Question

When silicon dioxide of thickness d is grown, the Silicon consumed is :

This question was previously asked in
UGC NET 2023 Electronic Science Question Paper (13-Dec-2023) (Shift 1)
The correct answer is

0.44 d

 Growing an oxide of thickness d consumes 0.44 d of silicon — option 2 — and the number can be derived rather than memorised.

The derivation. Thermal oxidation is

\(\text{Si}+\text{O}_{2}\rightarrow\text{SiO}_{2}\)

so each silicon atom consumed produces exactly one SiO2 molecule. The thickness ratio is therefore the ratio of the volume occupied per atom in the two materials:

 SiliconSiO2
Density2.33 g/cm32.21 g/cm3
Molar mass28.09 g/mol60.08 g/mol
Atoms or molecules per cm3\(5.0\times10^{22}\)\(2.2\times10^{22}\)

\(\dfrac{t_{Si}}{t_{ox}}=\dfrac{N_{ox}}{N_{Si}}=\dfrac{2.2\times10^{22}}{5.0\times10^{22}}=0.44\)

The oxide is less densely packed because each silicon atom now sits in a network with two oxygen atoms, so the same silicon spreads into more than twice the thickness.

What this means physically. Oxidation is not a coating laid on the surface — it consumes the wafer. Of the finished oxide, 44 % lies below the original surface and 56 % above it, so the interface moves into the silicon as growth proceeds. That is exactly why the Si-SiO2 interface is so clean: it is a freshly created interface buried beneath the oxide, never exposed to the atmosphere, which is the single most important reason silicon rather than germanium came to dominate.

Two practical consequences. First, in LOCOS isolation the growing field oxide sinks into the wafer and pushes up at the edges of the nitride mask, producing the characteristic "bird's beak" — a direct result of this 44 % consumption. Second, when a thin gate oxide is grown after a sacrificial oxide has been grown and stripped, the surface has already receded, and the process must account for it.

Growing 100 nm of oxide therefore consumes 44 nm of silicon, leaving the surface 56 nm above where it began.

Hence, the silicon consumed is 0.44 d.

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