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Musya8 [376]
3 years ago
7

In a metal-oxide-semiconductor (MOS) device, a thin layer of SiO2 (density = 2.20 Mg/m3) is grown on a single crystal chip of si

licon. How many Si atoms and how many O atoms are present per square millimeter of the oxide layer? Assume that the layer thickness is 160 nm.
Engineering
1 answer:
Troyanec [42]3 years ago
6 0

Answer:

3.52×10⁶ atoms of Si and 7.05×10⁶ atoms of O

Explanation:

It is all about unit conversions.

The area of our MOS is 1 mm². So, we know that the thickness is 160 nm. This data can give us the volume. We convert nm to mm.

160 nm . 1×10⁻⁶ mm /1nm = 1.6×10⁻⁴ mm

By the way, now we can determine the volume of MOS, in order to work with density.

1.6×10⁻⁴ mm . 1 mm² = 1.6×10⁻⁴ mm³

But density is mg/m³, so we convert mm³ to m³

1.6×10⁻⁴ mm³ . 1×10⁻⁹ m³/mm³ = 1.6×10⁻¹³ m³

Now, we apply density to determine the mass of MOS

Density = mass /volume → Density . volume = mass

1.6×10⁻¹³ m³ . 2.20mg/m³ = 3.52×10⁻¹³ mg

To make more easier the calculate, we convert mg to g.

3.52×10⁻¹³ mg . 1g /1000mg = 3.52×10⁻¹⁶ g

To count the atoms, we determine molar mass of SiO₂ → 60.08 g/mol

We need to know moles of Si and O₂ in the MOS

Firstly, we determine amount of MOS: 3.52×10⁻¹⁶ g / 60.08 g/mol = 5.86×10⁻¹⁸ moles

1 mol of SiO₂ has 1 mol of Si and 2 mol of O so:

5.86×10⁻¹⁸ mol of SiO₂ may have:

(5.86×10⁻¹⁸ . 1) /1 = 5.86×10⁻¹⁸ moles of Si

(5.86×10⁻¹⁸ .2) /1 =  1.17×10⁻¹⁷ moles of O₂

Let's count the atoms (1 mol of anything contain NA particles)

5.86×10⁻¹⁸ mol of Si . 6.02×10²³ atoms/ mol = 3.52×10⁶ atoms of Si

1.17×10⁻¹⁷ mol of O₂  . 6.02×10²³ atoms/ mol = 7.05×10⁶ atoms of O

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Explanation:

The catalog rating of a bearing can be found by using the following formula:

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