Atom count and dopant fraction in a silicon wafer: 22-Mec-B8 Engineering Materials, Undated paper, Q3

Question 3: Atom count and dopant fraction in a silicon wafer (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A standard 150 mm (six-inch) semiconductor wafer of pure single-crystal silicon:

Given data
Quantity	Symbol	Value
Wafer diameter	d	150 mm = 15.0 cm
Wafer thickness	t	0.5 mm = 0.05 cm
Density of silicon	ρ	2.33 g/cm³
Atomic mass of silicon	M	28.09 g/mol
Avogadro's number	N_A	6.022 × 10²³ mol⁻¹
Phosphorus doping level	N_P	10¹⁶ atoms/cm³

Find. (a) the total number of silicon atoms in the wafer, and (b) the atomic fraction of phosphorus once the wafer is doped.

The wafer, and the scale of the doping. The geometry gives 8.836 cm³ and hence 4.41 × 10²³ silicon atoms; the doping bar shows why the answer to (b) is so small — 10¹⁶ phosphorus atoms per cubic centimetre against 5.0 × 10²² silicon atoms is one part in five million, and the red line marking it has been drawn far wider than scale so that it is visible at all.

Approach. Get the volume from the wafer geometry, turn it into a mass and then into moles using the density and atomic mass, and multiply by Avogadro's number; for part (b), divide the stated phosphorus concentration by the silicon atom concentration the same calculation implies.

Part (a) — compute the volume of the wafer. Treating it as a right circular disc, $$V = \frac{\pi d^{2}}{4}t = \frac{\pi (15.0)^{2}}{4}(0.05) = \frac{\pi(225)(0.05)}{4}$$ $$V = 8.836\ \text{cm}^{3}$$ Real wafers carry a flat or a notch to mark the crystal orientation, which removes well under one per cent of the area; the question gives no such detail, so the full disc is used.
Convert the volume into a mass and then into moles. $$m = \rho V = (2.33)(8.836) = 20.59\ \text{g}$$ $$n = \frac{m}{M} = \frac{20.59}{28.09} = 0.7329\ \text{mol}$$ A six-inch wafer therefore holds about three quarters of a mole of silicon, which weighs a little over twenty grams — a useful figure to carry, because it makes the final answer easy to sanity-check.
Multiply by Avogadro's number. $$N = n N_A = (0.7329)(6.022\times10^{23})$$ $$\boxed{\ N = 4.41\times10^{23}\ \text{silicon atoms}\ }$$
Confirm the answer from the crystal structure instead of the density. Silicon has the diamond-cubic structure, eight atoms per unit cell, with $a = 0.5431$ nm, so $$N_{Si} = \frac{8}{(5.431\times10^{-8})^{3}} = 4.994\times10^{22}\ \text{atoms/cm}^{3}$$ and over the wafer volume that gives $(4.994\times10^{22})(8.836) = 4.412\times10^{23}$ atoms. The two routes agree to 0.03 per cent, which is a genuine check: one used the measured bulk density, the other the measured lattice parameter, and nothing but internal consistency links them.
Part (b) — form the atomic fraction of phosphorus. The silicon atom concentration follows from part (a) as $$N_{Si} = \frac{N}{V} = \frac{4.4135\times10^{23}}{8.836} = 5.00\times10^{22}\ \text{atoms/cm}^{3}$$ Phosphorus is a substitutional dopant, so it replaces silicon on the lattice and the atomic fraction is $$X_P = \frac{N_P}{N_P + N_{Si}} = \frac{10^{16}}{10^{16} + 5.00\times10^{22}}$$ $$\boxed{\ X_P = 2.00\times10^{-7}\ }$$
Restate the fraction in units an engineer will recognise. That is $2.00\times10^{-5}$ atomic per cent, or 0.20 atomic parts per million — one phosphorus atom for every five million silicon atoms. Because the dopant is so dilute, dropping it from the denominator changes nothing: $10^{16}/5.00\times10^{22}$ gives the same $2.00\times10^{-7}$ to three figures. The remarkable point is that a fraction this small transforms the material: intrinsic silicon at room temperature carries only about $1.5\times10^{10}$ carriers per cubic centimetre, so 10¹⁶ donors raise the free-electron concentration by roughly six orders of magnitude and the conductivity with it.

Results for the silicon wafer
Quantity	Relation	Value
Wafer volume	V = πd²t/4	8.836 cm³
Wafer mass	m = ρV	20.59 g
Moles of silicon	n = m/M	0.7329 mol
(a) Atoms in the wafer	N = nN_A	4.41 × 10²³ atoms
Check from the diamond-cubic cell	8/a³ × V	4.412 × 10²³ atoms (0.03 % apart)
Silicon atom concentration	N_Si = N/V	5.00 × 10²² cm⁻³
(b) Atomic fraction of phosphorus	X_P = N_P/(N_P+N_Si)	2.00 × 10⁻⁷ (0.20 atomic ppm)

Check: the density and atomic mass of silicon are not given in the question and are taken as 2.33 g/cm³ and 28.09 g/mol, the standard values; both should be quoted as assumptions in the answer book. The wafer is treated as a full disc, ignoring the orientation flat, which for a 150 mm wafer removes under one per cent of the volume. The doping is assumed uniform through the thickness, as the question's wording implies; a real wafer doped by ion implantation or by diffusion carries a strongly non-uniform profile, and 10¹⁶ cm⁻³ would then be a peak or an average rather than a constant.

Topic. Counting atoms from bulk properties, the atom density of a crystal structure, and the extreme dilution of semiconductor doping.

Key relations. N = ρVN_A/M; atom density from the unit cell, N = (atoms per cell)/a³, which is 8/a³ for the diamond-cubic structure of silicon, germanium and diamond; atomic fraction X = N_solute/(N_solute+N_solvent), reducing to N_solute/N_solvent when the solute is dilute; and for a doped semiconductor at room temperature the essentially complete ionisation of shallow donors, n ≈ N_D, with σ = nqμ_n.

Why this works. Two entirely independent measurements — the bulk density weighed on a balance, and the lattice parameter measured by X-ray diffraction — must give the same atom count, and the fact that they agree to a fraction of a per cent is one of the classical confirmations that a crystal really is a periodic array of discrete atoms. The doping calculation then makes the central point of semiconductor engineering: silicon holds about 5 × 10²² atoms per cubic centimetre, but at room temperature only about 1.5 × 10¹⁰ of them per cubic centimetre have thermally released an electron. Adding phosphorus at 0.2 atomic ppm injects 10¹⁶ donor electrons per cubic centimetre, six orders of magnitude more than the intrinsic supply, and the conductivity rises in proportion. This is why semiconductor-grade silicon must be purified to parts per billion before it is doped: an uncontrolled impurity at the level being deliberately added would swamp the design.

Common pitfall. Mixing millimetres and centimetres inside the volume calculation, which is out by a factor of 1000 for a linear-cubed quantity; using the radius where the formula wants the diameter, a factor of four; and, in part (b), dividing the doping level by the wafer's total atom count rather than by its atom concentration, which produces a dimensionally wrong answer some 24 orders of magnitude adrift.

Source. Askeland & Wright, The Science and Engineering of Materials, 7th ed., Ch. 3 (atomic and ionic arrangements; the diamond-cubic structure and atoms per cell) and Ch. 19 (electronic materials: doping and extrinsic semiconductors); Callister & Rethwisch, Ch. 3.5 and Ch. 18.11–18.12.

Question 3 of 8: Atom count and dopant fraction in a silicon wafer

Question 3: Atom count and dopant fraction in a silicon wafer (20 marks)