Ion engine — ionization, acceleration grid, neutralizer: Worked examples

Before any symbol appears in a worked example, here is the plain-English meaning of each, anchored to what it is physically:

Definition The cast of symbols

$q$ — the electric charge on one ion, measured in coulombs (C). A singly-charged ion has $q = + e = 1.602 \times 1 0^{- 19}$ C. Think of it as "how many handles the electric field has to grab this ion by."
$m$ — the mass of one ion in kilograms. Heavy ion = more momentum per push.
$V$ — the voltage (potential difference) the ion falls through, in volts. Picture a hill of height $V$ ; the ion rolls down and speeds up.
$v_{e}$ — the exhaust speed in m/s: how fast the ion leaves the back.
$I$ — the beam current in amperes (A): charge leaving per second. $1$ A $= 1$ C/s.
$\overset{m}{˙}$ — the mass flow rate in kg/s: how many kilograms of xenon leave per second.
$F$ — the thrust in newtons (N): the forward push on the ship.
$E_{i}$ — the ionization energy: the energy toll to rip one electron off, quoted in electron-volts (eV).

Prerequisites you may want open: Work-Energy Theorem, Specific Impulse and Exhaust Velocity, Tsiolkovsky Rocket Equation, Child-Langmuir Space Charge Law, Plasma Physics — Ionization Energy. This is a child of the ion-engine topic.

The scenario matrix

Every question this topic can ask lives in one of these cells. The last column names the example that clears it.

#	Case class	What changes / the trap	Worked in
A	Baseline singly-charged ion	plug straight in	Ex 1
B	Charge state $\neq = 1$ (doubly-charged Xe²⁺)	$q \to 2 e$ ; does $v_{e}$ double?	Ex 2
C	Zero / degenerate input ( $V = 0$ )	no acceleration → limiting value	Ex 3
D	Different propellant (light ion, e.g. argon)	mass changes, compare speeds	Ex 4
E	Thrust from current & voltage	use $F = I 2 mV / q$	Ex 5
F	Limiting behaviour (raise $V$ , space-charge cap)	why thrust does not scale like $v_{e}$	Ex 6
G	Neutralizer charge balance	electrons/sec, does it steal thrust?	Ex 7
H	Real-world word problem ( $Δ v$ over months)	combine thrust + time + Tsiolkovsky	Ex 8
I	Exam twist (energy accounting, eV)	ionization + power budget	Ex 9

Numbers reused throughout: $e = 1.602 \times 1 0^{- 19}$ C, $m_{Xe} = 2.18 \times 1 0^{- 25}$ kg, $1 eV = 1.602 \times 1 0^{- 19}$ J.

Ex 1 — Cell A: the baseline

Worked example Singly-charged xenon through

$V = 1500$ V. Find $v_{e}$ . Forecast: guess before computing — will it be nearer 4 km/s (chemical), 40 km/s, or 400 km/s?

Step 1. Identify the tool: the ion starts at rest and the field does work $q V$ on it. By the Work-Energy Theorem, all that work becomes kinetic energy: $q V = \frac{1}{2} m v_{e}^{2} .$ Why this step? No burning, no heat loss — the field is the only thing doing work, so 100% of $q V$ shows up as motion. That's the whole reason ion engines are fast.

Step 2. Solve for $v_{e}$ : $v_{e} = 2 q V / m$ . Why this step? We want speed, so isolate it — square-root because energy carries $v^{2}$ .

Step 3. Plug in: $v_{e} = \frac{2 ( 1.602 \times 1 0 ^{- 19} ) ( 1500 )}{2.18 \times 1 0 ^{- 25}} = 2.204 \times 1 0^{9} \approx 4.70 \times 1 0^{4} m/s .$

Verify: Units: $\frac{C \cdot V}{kg} = \frac{J}{kg} = \frac{m ^{2}}{s ^{2}}$ , square-rooted → m/s. ✓ Answer $\approx 47$ km/s sits between the parent's $42$ km/s (at $1200$ V) and higher — and $1500/1200 = 1.118$ , and $42 \times 1.118 = 47$ . ✓ Forecast: the $40$ km/s bracket was right.

Ex 2 — Cell B: doubly-charged ion (does $v_{e}$ double?)

Ex 3 — Cell C: the degenerate input $V = 0$

Ex 4 — Cell D: a lighter propellant (argon)

Worked example Swap xenon for

argon ( $m_{Ar} = 6.63 \times 1 0^{- 26}$ kg), singly-charged, same $V = 1500$ V. Compare $v_{e}$ to xenon. Forecast: argon is lighter — faster or slower exhaust?

Step 1. $v_{e} = 2 q V / m$ with $m = 6.63 \times 1 0^{- 26}$ : $v_{e} = \frac{2 ( 1.602 \times 1 0 ^{- 19} ) ( 1500 )}{6.63 \times 1 0 ^{- 26}} = 7.248 \times 1 0^{9} \approx 8.51 \times 1 0^{4} m/s .$

Step 2. Ratio to xenon (Ex 1): $v_{A r} / v_{X e} = m_{X e} / m_{A r} = 2.18 \times 1 0^{- 25} /6.63 \times 1 0^{- 26} = 3.288 = 1.813$ , and $47 \times 1.813 = 85$ km/s. ✓ Why this step? At fixed $V$ , $v_{e} \propto 1/ m$ — lighter ion, faster exhaust.

Step 3. But thrust per ion $= m v_{e} \propto m \cdot m^{- 1/2} = m$ : heavier ion wins on momentum. That is exactly why the parent note prefers heavy xenon despite argon's higher speed.

Verify: Units clean (m/s). $8.51/4.70 = 1.81 = 3.29$ . ✓ Forecast confirmed: lighter → faster, but not better for thrust.

Ex 5 — Cell E: thrust from current and voltage

Ex 6 — Cell F: the limiting twist — why thrust doesn't chase $v_{e}$

Figure — Ion engine — ionization, acceleration grid, neutralizer

Worked example Starting from

$V = 1500$ V, an engineer doubles the voltage to $3000$ V hoping to double thrust. What actually happens to $v_{e}$ and to the maximum extractable thrust? Forecast: does thrust double, rise by $2$ , or rise faster than $2$ ?

Step 1 — exhaust speed. $v_{e} \propto V$ , so doubling $V$ gives $v_{e} \times 2$ : $47 \to 66$ km/s. Why this step? Confirms efficiency (specific impulse) climbs. Look at the blue curve in the figure — it bends as a square root.

Step 2 — current ceiling. The grids can only pull so many ions per second before their own space charge blocks the flow. The Child-Langmuir Space Charge Law caps current as $I_{max} \propto V^{3/2}$ . Doubling $V$ raises the ceiling by $2^{3/2} = 2.83 \times$ . Why this step? Thrust needs both speed and ions; the ion supply is the real bottleneck.

Step 3 — thrust at the ceiling. $F = I 2 mV / q \propto V^{3/2} \cdot V^{1/2} = V^{2}$ only if you always run at the space-charge limit. Doubling $V$ then multiplies max thrust by $2^{2} = 4 \times$ — but this demands $4 \times$ the electrical power and grids that survive it. In practice power and grid erosion cap you, so you keep $I$ fixed and $F \propto V$ only (the red curve). Why this step? It resolves the parent's "higher voltage always means more thrust" myth: raw thrust does not scale simply — it depends on whether current or power is your limiter.

Verify: At fixed $I$ , $F \propto V$ : $F (3000) / F (1500) = 2 = 1.414$ . From Ex 5, $0.1125 \times 1.414 = 0.159$ N — plotted as the red point in the figure. ✓

Ex 7 — Cell G: neutralizer balance (does it steal thrust?)

Worked example The beam carries

$I_{beam} = 1.76$ A of Xe⁺. (a) How many electrons per second must the neutralizer emit? (b) How much backward momentum do those electrons carry versus the forward momentum of the ions? Forecast: will the electrons cancel a big fraction of the thrust, or almost none?

Step 1 — charge balance. Ions leave at $N = I / q = 1.76/1.602 \times 1 0^{- 19} = 1.098 \times 1 0^{19}$ per second. To keep the ship neutral, the neutralizer must emit the same number of electrons: $N_{e} = 1.098 \times 1 0^{19}$ /s, i.e. $I_{neut} = I_{beam} = 1.76$ A. Why this step? Charge in = charge out keeps the spacecraft at constant potential (parent's $I_{neut} = I_{beam}$ ).

Step 2 — momentum ratio. Electrons are emitted at low speed and are ~240,000× lighter than a Xe ion ( $m_{e} = 9.11 \times 1 0^{- 31}$ kg vs $2.18 \times 1 0^{- 25}$ kg). Per particle, at equal speed the momentum ratio would be $m_{e} / m_{Xe} = 4.18 \times 1 0^{- 6}$ . Electrons actually leave slower than ions, so their share of momentum is even tinier. Why this step? Thrust is momentum flow, not charge flow — the mass factor is what matters.

Verify: $m_{e} / m_{Xe} = 9.11 \times 1 0^{- 31} /2.18 \times 1 0^{- 25} = 4.18 \times 1 0^{- 6}$ — about four parts per million. The neutralizer removes essentially none of the thrust. ✓ Forecast (almost none) confirmed.

Ex 8 — Cell H: real-world $Δ v$ over months

Worked example A

$500$ kg probe runs the Ex 5 engine ( $F = 0.1125$ N, $\overset{m}{˙} = 2.395 \times 1 0^{- 6}$ kg/s) continuously for $6$ months ( $1.577 \times 1 0^{7}$ s). Ignoring the small mass loss, estimate the velocity change $Δ v$ . Then check with propellant mass used. Forecast: a $0.11$ N push for half a year — will $Δ v$ be a few m/s, or thousands?

Step 1 — impulse ≈ $F t$ . With nearly constant mass, $Δ v \approx \frac{F t}{M}$ . Why this step? Small thrust but huge time — impulse accumulates. This is the ion-engine trick.

Step 2 — numbers. $Δ v \approx \frac{0.1125 \times 1.577 \times 1 0 ^{7}}{500} = \frac{1.774 \times 1 0 ^{6}}{500} \approx 3549 m/s \approx 3.5 km/s .$

Step 3 — sanity via propellant. Xenon spent $= \overset{m}{˙} t = 2.395 \times 1 0^{- 6} \times 1.577 \times 1 0^{7} = 37.8$ kg. Using Tsiolkovsky Rocket Equation $Δ v = v_{e} ln \frac{M _{0}}{M _{f}}$ with $v_{e} = 4.70 \times 1 0^{4}$ m/s, $M_{0} = 500$ , $M_{f} = 500 - 37.8 = 462.2$ : $Δ v = 4.70 \times 1 0^{4} ln \frac{500}{462.2} = 4.70 \times 1 0^{4} \times 0.07858 \approx 3693 m/s .$ Why this step? The rocket equation is the honest bookkeeping; the $F t / M$ estimate ignores mass loss so reads a touch low.

Verify: Two methods give $3.5$ – $3.7$ km/s — the same ballpark, the difference exactly the mass-loss correction. ✓ Forecast (thousands of m/s) confirmed: patience beats brute force.

Ex 9 — Cell I: exam twist — energy accounting

Worked example For every ion the engine makes, energy is spent

twice: once to ionize ( $E_{i} = 12.13$ eV) and once to accelerate (through $V = 1500$ V, so $q V$ corresponds to $1500$ eV per singly-charged ion). What fraction of the electrical energy per ion is "wasted" on ionization rather than turned into exhaust kinetic energy? Forecast: is ionization a big overhead (tens of percent) or a small tax (a few percent)?

Step 1 — measure both in eV. A singly-charged ion falling through $1500$ V gains $1500$ eV of kinetic energy. Ionizing it cost $12.13$ eV. Why this step? eV is the natural unit: "energy an electron gains crossing $1$ V." Same currency for both terms means no unit juggling.

Step 2 — fraction. $wasted fraction = \frac{E _{i}}{E _{i} + q V _{eV}} = \frac{12.13}{12.13 + 1500} = \frac{12.13}{1512.13} \approx 0.00802.$

Step 3 — read it. About $0.80%$ of the per-ion energy goes to ionization; over $99%$ becomes exhaust kinetic energy (ideal case). Ionization is a small tax. Why this step? Explains why ion engines are so efficient: the expensive-sounding "rip an electron off" step is cheap compared with the acceleration.

Verify: $12.13/1512.13 = 0.008022$ → $0.80%$ . ✓ Forecast (few percent) confirmed.

Recall Scenario checklist — can you place each in the matrix?

Doubling charge multiplies $v_{e}$ by ::: $2$ (since $v_{e} \propto q$ ). $V = 0$ gives exhaust speed and thrust equal to ::: zero (no downhill to fall). Lighter ion (argon) at fixed $V$ gives a exhaust speed that is ::: faster ( $v_{e} \propto 1/ m$ ), but lower momentum per ion. At the space-charge ceiling, max thrust scales like ::: $V^{2}$ ; at fixed current only like $V$ . Neutralizer electrons steal roughly what fraction of thrust ::: about $4$ parts per million — negligible. A $0.11$ N engine for 6 months on a $500$ kg probe gives $Δ v \approx$ ::: $3.5$ – $3.7$ km/s. Ionization overhead at $1500$ V is about ::: $0.8%$ of per-ion energy.