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

3.3.41 · D3 · Physics › Rocket Propulsion › Ion engine — ionization, acceleration grid, neutralizer

Kisi bhi worked example mein symbol aane se pehle, yahan har ek ka plain-English meaning diya gaya hai — anchored to that it physically kya hai:

Definition Symbols ki cast

$q$ — ek ion par electric charge, coulombs (C) mein measure hota hai. Ek singly-charged ion ka $q = + e = 1.602 \times 1 0^{- 19}$ C hota hai. Socho aise — "electric field ke paas is ion ko pakadne ke kitne handle hain."
$m$ — ek ion ki mass kilograms mein. Bhaari ion = ek push mein zyada momentum.
$V$ — ion jis voltage (potential difference) se girta hai, volts mein. Ek pahaadi socho jis ki height $V$ hai; ion neeche roll karta hai aur speed pakad leta hai.
$v_{e}$ — exhaust speed m/s mein: ion kitni tezi se peeche se nikalta hai.
$I$ — beam current amperes (A) mein: charge jo per second nikal raha hai. $1$ A $= 1$ C/s.
$\overset{m}{˙}$ — mass flow rate kg/s mein: per second kitne kilograms xenon nikal raha hai.
$F$ — thrust newtons (N) mein: ship par aage ki taraf push.
$E_{i}$ — ionization energy: ek electron ko todne ki energy cost, electron-volts (eV) mein batayi jaati hai.

Prerequisites jo tum open rakhna chahoge: Work-Energy Theorem, Specific Impulse and Exhaust Velocity, Tsiolkovsky Rocket Equation, Child-Langmuir Space Charge Law, Plasma Physics — Ionization Energy. Ye ion-engine topic ka ek child hai.

The scenario matrix

Is topic se jo bhi question aa sakta hai, woh in cells mein se kisi ek mein hoga. Last column mein us example ka naam hai jo use clear karta hai.

#	Case class	Kya badlta hai / trap kya hai	Worked in
A	Baseline singly-charged ion	seedha plug in karo	Ex 1
B	Charge state $\neq = 1$ (doubly-charged Xe²⁺)	$q \to 2 e$ ; kya $v_{e}$ double hoga?	Ex 2
C	Zero / degenerate input ( $V = 0$ )	no acceleration → limiting value	Ex 3
D	Alag propellant (light ion, e.g. argon)	mass badlti hai, speeds compare karo	Ex 4
E	Current aur voltage se Thrust	$F = I 2 mV / q$ use karo	Ex 5
F	Limiting behaviour ( $V$ badhao, space-charge cap)	thrust kyun nahi scale karta $v_{e}$ jaisa	Ex 6
G	Neutralizer charge balance	electrons/sec, kya ye thrust chura leta hai?	Ex 7
H	Real-world word problem ( $Δ v$ mahino mein)	thrust + time + Tsiolkovsky combine karo	Ex 8
I	Exam twist (energy accounting, eV)	ionization + power budget	Ex 9

Poore note mein use hone wale numbers: $e = 1.602 \times 1 0^{- 19}$ 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

$V = 1500$ V se guzra. $v_{e}$ nikalo. Forecast: compute karne se pehle andaza lagao — kya ye 4 km/s (chemical) ke paas hoga, 40 km/s, ya 400 km/s?

Step 1. Tool identify karo: ion rest se start karta hai aur field usse $q V$ kaam karta hai. Work-Energy Theorem ke by, woh saara kaam kinetic energy ban jaata hai: $q V = \frac{1}{2} m v_{e}^{2} .$ Ye step kyun? Koi burning nahi, koi heat loss nahi — field akela kaam kar raha hai, isliye $q V$ ka 100% motion ban jaata hai. Yahi wajah hai ki ion engines itni fast hoti hain.

Step 2. $v_{e}$ ke liye solve karo: $v_{e} = 2 q V / m$ . Ye step kyun? Humein speed chahiye, toh ise isolate karo — square-root isliye kyunki energy mein $v^{2}$ hota hai.

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

Verify: Units: $\frac{C \cdot V}{kg} = \frac{J}{kg} = \frac{m ^{2}}{s ^{2}}$ , square-root liya → m/s. ✓ Answer $\approx 47$ km/s, parent ke $42$ km/s ( $1200$ V par) aur zyada ke beech mein baithta hai — aur $1500/1200 = 1.118$ , aur $42 \times 1.118 = 47$ . ✓ Forecast: $40$ km/s bracket sahi tha.

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

Worked example Same xenon, lekin

do electrons tod diye → Xe²⁺, toh $q = 2 e$ , same $V = 1500$ V. $v_{e}$ nikalo. Forecast: double charge — kya tum $2 \times$ speed, $2 \times$ , ya unchanged expect karte ho?

Step 1. Same equation, naya $q = 2 e = 3.204 \times 1 0^{- 19}$ C. Ye step kyun? Charge state sirf $q$ ke through enter karta hai; Xe²⁺ ki mass essentially unchanged hai (humne sirf do feather-light electrons nikale hain).

Step 2. $v_{e} = \frac{2 ( 3.204 \times 1 0 ^{- 19} ) ( 1500 )}{2.18 \times 1 0 ^{- 25}} = 4.408 \times 1 0^{9} \approx 6.64 \times 1 0^{4} m/s .$

Step 3. Ex 1 se compare karo: ratio $= 2 \approx 1.414$ , aur $47 \times 1.414 = 66.4$ km/s. ✓ Ye step kyun? $v_{e} \propto q$ hai, $q$ nahi. Charge double karne se speed $2$ se multiply hoti hai, $2$ se nahi.

Verify: $6.64/4.70 = 1.413 \approx 2$ . ✓ Matrix cell ke liye takeaway: charge state matter karta hai, lekin square root ke andar.

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

Ex 4 — Cell D: lighter propellant (argon)

Worked example Xenon ki jagah

argon ( $m_{Ar} = 6.63 \times 1 0^{- 26}$ kg) use karo, singly-charged, same $V = 1500$ V. $v_{e}$ ko xenon se compare karo. Forecast: argon halka hai — exhaust faster hoga ya slower?

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

Step 2. Xenon (Ex 1) se ratio: $v_{A r} / v_{X e} = m_{X e} / m_{A r} = 2.18 \times 1 0^{- 25} /6.63 \times 1 0^{- 26} = 3.288 = 1.813$ , aur $47 \times 1.813 = 85$ km/s. ✓ Ye step kyun? Fixed $V$ par, $v_{e} \propto 1/ m$ — halka ion, faster exhaust.

Step 3. Lekin thrust per ion $= m v_{e} \propto m \cdot m^{- 1/2} = m$ : bhaari ion momentum par jeet jaata hai. Yahi exact wajah hai ki parent note heavy xenon ko prefer karta hai argon ki higher speed ke bawajood.

Verify: Units clean hain (m/s). $8.51/4.70 = 1.81 = 3.29$ . ✓ Forecast confirm hua: halka → faster, lekin thrust ke liye better nahi.

Ex 5 — Cell E: current aur voltage se thrust

Worked example Ek grid

$V = 1500$ V par Xe⁺ ka beam current $I = 1.76$ A draw karta hai. Thrust $F$ nikalo. Forecast: coin-weight ( $\sim 0.1$ N), gram-weight, ya newton-scale?

Step 1. Thrust ka current form choose karo: $F = I \frac{2 mV}{q} .$ Ye step kyun? Humein current diya gaya hai, mass flow directly nahi. Kyunki charge ka har coulomb = $1/ q$ ions jinka har ek ka mass $m$ hai, current already $\overset{m}{˙}$ encode kar leta hai; ye form ek conversion bachata hai.

Step 2. Root ke andar: $\frac{2 ( 2.18 \times 1 0 ^{- 25} ) ( 1500 )}{1.602 \times 1 0 ^{- 19}} = \frac{6.54 \times 1 0 ^{- 22}}{1.602 \times 1 0 ^{- 19}} = 4.082 \times 1 0^{- 3}$ ; root $= 0.0639$ .

Step 3. $F = 1.76 \times 0.0639 \approx 0.1125 N$ .

Verify: $F = \overset{m}{˙} v_{e}$ se cross-check karo. Mass flow $\overset{m}{˙} = (I / q) m = (1.76/1.602 \times 1 0^{- 19}) (2.18 \times 1 0^{- 25}) = 2.395 \times 1 0^{- 6}$ kg/s. Phir $\overset{m}{˙} v_{e} = 2.395 \times 1 0^{- 6} \times 4.70 \times 1 0^{4} = 0.1126$ N. ✓ Do independent routes agree karte hain; forecast (coin-scale) confirm hua.

Ex 6 — Cell F: limiting twist — thrust $v_{e}$ ke peeche kyun nahi bhaagta

Figure — Ion engine — ionization, acceleration grid, neutralizer

Worked example

$V = 1500$ V se shuru hokar, ek engineer voltage double karta hai $3000$ V par, thrust double karne ki umeed mein. $v_{e}$ aur extractable thrust ke maximum ke saath actually kya hota hai? Forecast: kya thrust double hoga, $2$ se badhega, ya $2$ se zyada tezi se badhega?

Step 1 — exhaust speed. $v_{e} \propto V$ , toh $V$ double karne par $v_{e} \times 2$ milta hai: $47 \to 66$ km/s. Ye step kyun? Confirm karta hai ki efficiency (specific impulse) badhti hai. Figure mein blue curve dekho — woh square root ki tarah bend karti hai.

Step 2 — current ceiling. Grids sirf utne hi ions per second pull kar sakti hain jab tak unka apna space charge flow block na kar de. Child-Langmuir Space Charge Law current ko cap karta hai $I_{max} \propto V^{3/2}$ se. $V$ double karne par ceiling $2^{3/2} = 2.83 \times$ badhti hai. Ye step kyun? Thrust ke liye dono speed aur ions chahiye; ion supply hi asli bottleneck hai.

Step 3 — ceiling par thrust. $F = I 2 mV / q \propto V^{3/2} \cdot V^{1/2} = V^{2}$ sirf tab jab tum hamesha space-charge limit par run karo. Tab $V$ double karne par max thrust $2^{2} = 4 \times$ multiply hota hai — lekin iske liye $4 \times$ electrical power aur grids chahiye jo ise survive kar sakein. Practice mein power aur grid erosion tumhe cap kar deti hai, toh tum $I$ fixed rakhte ho aur $F \propto V$ sirf (red curve). Ye step kyun? Ye parent ki "higher voltage matlab hamesha more thrust" myth resolve karta hai: raw thrust simply scale nahi karta — ye depend karta hai ki current ya power tumhari limiter hai.

Verify: Fixed $I$ par, $F \propto V$ : $F (3000) / F (1500) = 2 = 1.414$ . Ex 5 se, $0.1125 \times 1.414 = 0.159$ N — figure mein red point ki tarah plot hua. ✓

Ex 7 — Cell G: neutralizer balance (kya ye thrust chura leta hai?)

Worked example Beam

$I_{beam} = 1.76$ A Xe⁺ carry karta hai. (a) Neutralizer ko per second kitne electrons emit karne chahiye? (b) Un electrons ka backward momentum ions ke forward momentum ke mukable mein kitna hai? Forecast: kya electrons thrust ka bada fraction cancel kar denge, ya almost kuch nahi?

Step 1 — charge balance. Ions $N = I / q = 1.76/1.602 \times 1 0^{- 19} = 1.098 \times 1 0^{19}$ per second nikalte hain. Ship ko neutral rakhne ke liye, neutralizer ko same number electrons emit karne chahiye: $N_{e} = 1.098 \times 1 0^{19}$ /s, yaani $I_{neut} = I_{beam} = 1.76$ A. Ye step kyun? Charge in = charge out spacecraft ko constant potential par rakhta hai (parent ka $I_{neut} = I_{beam}$ ).

Step 2 — momentum ratio. Electrons kam speed par emit hote hain aur Xe ion se ~240,000× halke hote hain ( $m_{e} = 9.11 \times 1 0^{- 31}$ kg vs $2.18 \times 1 0^{- 25}$ kg). Per particle, equal speed par momentum ratio hoga $m_{e} / m_{Xe} = 4.18 \times 1 0^{- 6}$ . Electrons actually ions se slower nikalte hain, toh unka momentum share aur bhi tiny hai. Ye step kyun? Thrust momentum flow hai, charge flow nahi — mass factor hi matter karta hai.

Verify: $m_{e} / m_{Xe} = 9.11 \times 1 0^{- 31} /2.18 \times 1 0^{- 25} = 4.18 \times 1 0^{- 6}$ — lagbhag chaar parts per million. Neutralizer essentially koi thrust nahi churaata. ✓ Forecast (almost none) confirm hua.

Ex 8 — Cell H: real-world $Δ v$ mahino mein

Worked example Ek

$500$ kg probe Ex 5 wala engine ( $F = 0.1125$ N, $\overset{m}{˙} = 2.395 \times 1 0^{- 6}$ kg/s) lagaataar $6$ months ( $1.577 \times 1 0^{7}$ s) run karta hai. Chhote mass loss ko ignore karte hue, velocity change $Δ v$ estimate karo. Phir use propellant mass se check karo. Forecast: $0.11$ N ki push aadhe saal ke liye — kya $Δ v$ kuch m/s hoga, ya hajaaron?

Step 1 — impulse $\approx F t$ . Nearly constant mass ke saath, $Δ v \approx \frac{F t}{M}$ . Ye step kyun? Choti thrust lekin huge time — impulse accumulate hoti hai. Yahi ion-engine ki trick hai.

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 — propellant se sanity check. Xenon kharch hua $= \overset{m}{˙} t = 2.395 \times 1 0^{- 6} \times 1.577 \times 1 0^{7} = 37.8$ kg. Tsiolkovsky Rocket Equation $Δ v = v_{e} ln \frac{M _{0}}{M _{f}}$ use karo $v_{e} = 4.70 \times 1 0^{4}$ m/s, $M_{0} = 500$ , $M_{f} = 500 - 37.8 = 462.2$ ke saath: $Δ v = 4.70 \times 1 0^{4} ln \frac{500}{462.2} = 4.70 \times 1 0^{4} \times 0.07858 \approx 3693 m/s .$ Ye step kyun? Rocket equation honest bookkeeping hai; $F t / M$ estimate mass loss ignore karta hai isliye thoda kam read karta hai.

Verify: Do methods $3.5$ – $3.7$ km/s dete hain — same ballpark, difference exactly mass-loss correction hai. ✓ Forecast (hajaaron m/s) confirm hua: patience brute force ko haraata hai.

Ex 9 — Cell I: exam twist — energy accounting

Worked example Engine jo bhi ion banata hai, uske liye energy

do baar kharch hoti hai: ek baar ionize karne mein ( $E_{i} = 12.13$ eV) aur ek baar accelerate karne mein ( $V = 1500$ V se, toh $q V$ corresponds karta hai singly-charged ion ke liye $1500$ eV per ion). Per ion electrical energy ka kitna fraction ionization par "waste" hota hai, exhaust kinetic energy mein badle jaane ki bajaye? Forecast: kya ionization ek badi overhead hai (tens of percent) ya ek chhota sa tax (kuch percent)?

Step 1 — dono ko eV mein measure karo. Ek singly-charged ion $1500$ V se girne par $1500$ eV kinetic energy gain karta hai. Ise ionize karne mein $12.13$ eV laga. Ye step kyun? eV natural unit hai: "ek electron ko $1$ V cross karne par jo energy milti hai." Dono terms ke liye same currency matlab koi unit juggling nahi.

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 — padho ise. Per-ion energy ka lagbhag $0.80%$ ionization mein jaata hai; $99%$ se zyada exhaust kinetic energy ban jaata hai (ideal case mein). Ionization ek chota sa tax hai. Ye step kyun? Ye explain karta hai ki ion engines itni efficient kyun hain: "ek electron todo" wala step, jo expensive-sounding lagta hai, acceleration ke comparison mein sasta hai.

Verify: $12.13/1512.13 = 0.008022$ → $0.80%$ . ✓ Forecast (kuch percent) confirm hua.

Recall Scenario checklist — kya tum har ek ko matrix mein place kar sakte ho?

Charge double karne par $v_{e}$ ::: $2$ se multiply hoti hai (kyunki $v_{e} \propto q$ ). $V = 0$ par exhaust speed aur thrust dono barabar hote hain ::: zero (girne ke liye koi downhill nahi). Lighter ion (argon) fixed $V$ par exhaust speed deta hai jo ::: faster hoti hai ( $v_{e} \propto 1/ m$ ), lekin per ion momentum kam hota hai. Space-charge ceiling par, max thrust scale karta hai ::: $V^{2}$ ki tarah; fixed current par sirf $V$ ki tarah. Neutralizer electrons thrust ka roughly kitna fraction churaate hain ::: lagbhag $4$ parts per million — negligible. $500$ kg probe par 6 months ke liye $0.11$ N engine $Δ v \approx$ deta hai ::: $3.5$ – $3.7$ km/s. $1500$ V par ionization overhead lagbhag hai ::: per-ion energy ka $0.8%$ .