Question bank — Ion engine — ionization, acceleration grid, neutralizer

False. It's chosen for being heavy (big momentum per ion), easy to ionize (12.13 eV), inert and storable — xenon is actually rare and expensive.

False. Electric fields only push charged objects; a neutral atom feels no force, which is exactly why Stage 1 (ionization) must come first.

False. Thrust comes from momentum $mv$, not charge. The ion is ~240,000× heavier than an electron, so it carries essentially all the momentum; electrons only fix charge.

False. Since $v_e=\sqrt{2qV/m}$, exhaust speed scales with $\sqrt{V}$, so doubling $V$ multiplies $v_e$ by only $\sqrt{2}\approx 1.41$.

True. Thrust is tiny (≈0.1 N, coin-weight) because ion flow rate is small — but it runs for months, giving huge total $\Delta v$. See Chemical vs Electric Propulsion.

True. If $I_{neutralizer}=I_{beam}$, charge leaving as ions is matched by charge leaving as electrons, so the spacecraft stays at constant potential.

False. Xenon is the reaction mass that's thrown out; electricity only accelerates it. Run out of xenon → no thrust, even with full power.

False. Thrust is current-limited by how many ions can be extracted (Child–Langmuir limit); extra $V$ mainly buys efficiency (specific impulse), not big thrust. See Child-Langmuir Space Charge Law.

False. That freed electron goes on to ionize more atoms, sustaining the discharge — it's a self-feeding chain, not waste.

Error: electrons have negligible mass (~1/240000 of a xenon ion) and low speed, so their contribution to $\dot{m}{v}_e$ is essentially zero.

Error: light ions go faster but carry less momentum. Thrust $F=I\sqrt{2mV/q}$ actually rises with mass $m$, favouring heavy ions.

Error: only the magnitude matters. A positive ion falls through the positive-to-negative drop; the work $qV$ is positive because the field points the ion the right way.

Error: the screen grid is high positive; it's the outer accelerator grid that is strongly negative, pulling positive ions outward and blocking electrons from flowing back in.

Error: both leave the same direction (out the back); and even so, momentum is $mv$, dominated entirely by the massive ions.

Error: that's the ideal single-ion picture. Real engines lose power to ionization cost, grid interception, and imperfect neutralization.

Because the grids act via an electric field, and a neutral atom feels no electric force — you must first make it charged so the field can grab it.

Each ionizing collision produces one ion plus an extra free electron; that electron ionizes further atoms, so one seed electron triggers a chain.

It ejects positive ions continuously; losing positive charge each second leaves the ship with a growing net negative charge.

A negative ship electrostatically pulls the departing positive ions back, cancelling the forward momentum they were meant to carry.

Their thrust (~0.1 N) is far below the ship's Earth weight, so they can't lift off; but in space, months of tiny push add up to enormous $\Delta v$. See Tsiolkovsky Rocket Equation.

Thrust is momentum flow $\dot{m}{v}_e$; the heavier mass more than compensates for the slower speed, since $F=I\sqrt{2mV/q}$ grows with $m$.

It directly equates the electrical work $qV$ done on the ion to its gained kinetic energy $\frac{1}{2}m{v}_e^2$, needing no details of the field's shape. See Work-Energy Theorem.

Specific impulse captures fuel economy (thrust per propellant used); ion engines score high there even while their raw thrust stays tiny. See Specific Impulse and Exhaust Velocity.

Because a wider gap $d$ lets the beam's own positive space charge build up more, choking the current ($I\propto 1/d^2$); the gap sets how crowded the ions become. See Child-Langmuir Space Charge Law.

Thrust vanishes: $F=I\sqrt{2mV/q}=0$. Voltage alone with no ions moving produces no momentum flow.

The spacecraft charges positive, which would then attract electrons back and repel nothing useful — the balance $I_{neutralizer}=I_{beam}$ must hold both ways.

Each ion gains more energy ($v_e\propto \sqrt{q}$), but per unit current fewer ions carry the charge; doubly-charged ions are usually an inefficiency, wasting ionization energy without matching thrust gain.

No. Even ignoring hardware limits, thrust is capped by the space-charge (Child–Langmuir) limit on extractable current; $v_e$ keeps rising but $I$ can't keep pace. See Child-Langmuir Space Charge Law.

Same voltage gives much higher $v_e$ but far less momentum per ion, so thrust drops — this is why heavy inert xenon is preferred. See Plasma Physics — Ionization Energy.

With $I\propto V^{3/2}/d^2$ the extractable current blows up unphysically and the grids would arc/short — real engines need a finite gap, showing the geometry limit is fundamental.