Physical vapor deposition (PVD - sputtering): Worked examples

This page is a problem gym for PVD sputtering. The parent note gave you three formulas. Here we push each one into every corner it can reach — light ions, heavy ions, equal masses, degenerate zero-energy cases, real chamber numbers, and an exam-style twist.

Before anything, let's restate the three tools with plain-word meaning, because we will use their symbols relentlessly below.

The scenario matrix

Every problem this topic can throw is one of these cells. The worked examples below are tagged with the cell they cover.

Cell	What varies	Physical question	Example
A	$M_{1} \approx M_{2}$	Best-case transfer	Ex 1
B	$M_{1} ≪ M_{2}$ (light ion, heavy target)	Poor transfer	Ex 2
C	$M_{1} ≫ M_{2}$ (heavy ion, light target)	Symmetry check	Ex 3
D	Threshold across materials	Which needs more energy?	Ex 4
E	Degenerate: $E \to 0$ or $E < E_{t h}$	Nothing ejected	Ex 5
F	Limiting: $M_{1} = M_{2}$ exactly	$γ$ maxes at 1	Ex 6
G	Real-world word problem	Rate & time from flux	Ex 7
H	Exam twist: choose gas to maximize $γ$	Optimization	Ex 8

Notice the symmetry hiding in $γ$ : swap $M_{1} \leftrightarrow M_{2}$ and the formula is unchanged (numerator $4 M_{1} M_{2}$ and denominator $(M_{1} + M_{2})^{2}$ are both symmetric). Cells B and C should therefore give the same $γ$ for swapped masses — Ex 2 and Ex 3 are designed to prove that.

Figure — Physical vapor deposition (PVD - sputtering)

Look at the red curve above: $γ$ plotted against the mass ratio $M_{1} / M_{2}$ . It peaks at exactly 1 and falls off symmetrically on a log axis. Every example below is a single point on this one curve — keep glancing back at it.

Cell A — masses nearly equal (efficient transfer)

Cell B — light ion, heavy target (poor transfer)

Cell C — heavy ion, light target (symmetry proof)

Cell D — threshold across different materials

Worked example Ex 4 · Threshold energy: Al vs W with Ar ions

Surface binding energies: aluminum $U_{s} = 3.4$ eV, tungsten $U_{s} = 8.9$ eV. Ion is argon ( $M_{1} = 40$ ). Aluminum $M_{2} = 27$ , tungsten $M_{2} = 184$ . Find each $E_{t h}$ .

Forecast: tungsten is heavier and more tightly bound — its threshold should be much larger.

Al: reuse $γ = 0.962$ (Ex 1). $E_{t h} = \frac{3.4}{0.962} = 3.53$ eV. Why this step? $E_{t h} = U_{s} / γ$ — divide the binding energy by the fraction that actually gets delivered. Small $U_{s}$ , big $γ$ ⇒ tiny threshold.
W: compute its $γ$ first. $γ = \frac{4 ( 40 ) ( 184 )}{( 40 + 184 ) ^{2}} = \frac{29440}{50176} = 0.587$ . Why this step? Tungsten is a different point on the curve — we cannot reuse aluminum's $γ$ .
W threshold. $E_{t h} = \frac{8.9}{0.587} = 15.2$ eV. Why this step? Both effects (bigger $U_{s}$ , smaller $γ$ ) push the threshold up — the formula multiplies them together via the divide.

Verify: $15.2 > 3.53$ ✓ tungsten's threshold is ~4× aluminum's. In practice we run ions at hundreds of eV, so both are easily above threshold — this is why sputtering handles even refractory metals.

Cell E — degenerate inputs (nothing happens)

Worked example Ex 5 · Below threshold and zero energy

An argon ion hits aluminum with (a) $E = 2$ eV, (b) $E = 0$ eV. How many atoms eject?

Forecast: Ex 4 told us aluminum needs 3.53 eV. Both 2 and 0 are below it — guess zero yield.

Energy actually delivered at 2 eV. $E_{t r an s f er} = γ E = 0.962 \times 2 = 1.92$ eV. Why this step? The atom only escapes if what it receives beats its binding $U_{s} = 3.4$ eV.
Compare. $1.92 < 3.4$ ⇒ 0 atoms ejected. Why this step? This is the physical meaning of a threshold: below it, the yield $Y$ is exactly zero, not "a little bit".
Zero-energy case. $E = 0 \Rightarrow E_{t r an s f er} = 0 \Rightarrow$ 0 atoms. Why this step? Degenerate check — a stationary ion transfers nothing. Sputtering literally cannot start; this is why a minimum voltage is required to strike a useful discharge.

Verify: both give $Y = 0$ ✓. Units check: eV compared to eV ✓. The plasma would deposit no film in either case.

Cell F — the exact peak ( $M_{1} = M_{2}$ )

Cell G — real-world word problem (rate & time)

Worked example Ex 7 · How long to deposit a 100 nm aluminum film?

A magnetron delivers ion flux $J = 3.0 \times 1 0^{16}$ ions·cm⁻²·s⁻¹ at yield $Y = 1.2$ atoms/ion. Aluminum has $N = 6.0 \times 1 0^{22}$ atoms·cm⁻³ that must be packed to make the film. How long to grow $d = 100$ nm $= 1.0 \times 1 0^{- 5}$ cm?

Forecast: flux is huge, so guess seconds, not minutes.

Atom arrival rate per area. $Φ = Y \cdot J = 1.2 \times 3.0 \times 1 0^{16} = 3.6 \times 1 0^{16}$ atoms·cm⁻²·s⁻¹. Why this step? $R \propto Y \cdot J$ from the parent note — each ion ejects $Y$ atoms, and $J$ ions arrive per area per second. Multiply them.
Atoms needed per area for thickness $d$ . $N \cdot d = 6.0 \times 1 0^{22} \times 1.0 \times 1 0^{- 5} = 6.0 \times 1 0^{17}$ atoms·cm⁻². Why this step? A film of thickness $d$ over 1 cm² is a box of volume $d \times 1 cm^{2}$ ; multiply by atom density $N$ to get atoms.
Time = need ÷ rate. $t = \frac{6.0 \times 1 0 ^{17}}{3.6 \times 1 0 ^{16}} = 16.7$ s. Why this step? (atoms needed) ÷ (atoms arriving per second) = seconds. Units cancel cleanly: $\frac{atoms/cm ^{2}}{atoms/cm ^{2} /s} = s$ .

Verify: ~17 s ✓ — order-of-magnitude sensible for a fast magnetron. Units: cm²·s⁻¹ terms cancel to seconds ✓.

Cell H — exam twist (pick the best gas)

Worked example Ex 8 · Which noble gas maximizes

$γ$ for a copper target? Copper $M_{2} = 63.5$ . Candidate ions: neon (20), argon (40), krypton (84), xenon (131). Which gives the largest $γ$ ?

Forecast: the peak is at $M_{1} = M_{2} = 63.5$ . Krypton (84) is closest to 63.5 — guess krypton wins.

Compute $γ$ for each using $γ = \frac{4 M _{1} ( 63.5 )}{( M _{1} + 63.5 ) ^{2}}$ :
- Ne: $\frac{4 ( 20 ) ( 63.5 )}{( 83.5 ) ^{2}} = \frac{5080}{6972.25} = 0.729$
- Ar: $\frac{4 ( 40 ) ( 63.5 )}{( 103.5 ) ^{2}} = \frac{10160}{10712.25} = 0.948$
- Kr: $\frac{4 ( 84 ) ( 63.5 )}{( 147.5 ) ^{2}} = \frac{21336}{21756.25} = 0.981$
- Xe: $\frac{4 ( 131 ) ( 63.5 )}{( 194.5 ) ^{2}} = \frac{33274}{37830.25} = 0.880$ Why this step? We evaluate the same curve at four mass-ratio points to find the maximum — this is the whole matrix collapsed into one comparison.
Pick the biggest. Krypton, $γ = 0.981$ . Why this step? It sits nearest the mass-match peak, so it hands over the most energy per hit.

Verify: ordering Ne $<$ Xe $<$ Ar $<$ Kr in $γ$ ✓, with Kr on top ✓. In real fabs argon is still the default despite krypton's edge — argon is cheaper and lighter to pump, and the extra few percent of $γ$ rarely justifies expensive krypton. (Great exam trap: the "physics-best" answer isn't always the "engineering" answer.)

Recall Self-test (reveal after guessing)

$γ$ for Ar (40) on Cu (63.5) ::: $0.948$ Threshold energy for aluminum with argon ( $U_{s} = 3.4$ , $γ = 0.962$ ) ::: $\approx 3.53$ eV Yield when delivered energy is below $U_{s}$ ::: exactly $0$ atoms Time to grow 100 nm Al at $Φ = 3.6 \times 1 0^{16}$ atoms·cm⁻²·s⁻¹, $N = 6 \times 1 0^{22}$ ::: $\approx 16.7$ s Best noble gas for a copper target (physics answer) ::: krypton, $γ \approx 0.981$

See also: Mean Free Path, Plasma Physics, Thin Film Deposition, Interconnect Metallization.