Momentum and voltage: a simple link between a charged particle’s momentum and the accelerating potential

Imagine a tiny electron, a speck of charge, zooming through vacuum as if it’s on rails—driven by voltage and guided by fields. The big question? How do we quantify its motion in a compact, universal way. That answer pops out cleanly in the momentum relation p = sqrt(2 e m V). Yes, that little formula is doing a lot of heavy lifting.

Let’s connect the dots, step by step, and keep it practical.

From kinetic energy to momentum: the heart of the idea

When a charged particle is pulled through a potential difference V, the electric field does work on it. The work done on the particle turns into kinetic energy. For a particle with charge e, that gained energy is simply eV. In symbols: K.E. = eV.

But momentum and kinetic energy aren’t the same thing. Kinetic energy is also p^2/(2m) for a particle of mass m moving with momentum p. So we have two expressions for the same energy:

• K.E. = eV

• K.E. = p^2 / (2m)

Set them equal to each other and solve for p:

eV = p^2 / (2m)

p^2 = 2 m e V

p = sqrt(2 m e V)

That’s the momentum in terms of charge, mass, and the accelerating voltage. Quick, elegant, and very NEET-friendly because it connects a voltage to a motion parameter.

A tiny numerical moment to anchor things

The numbers help. Take an electron (e ≈ 1.6 × 10^-19 C, m ≈ 9.11 × 10^-31 kg). Suppose it’s accelerated through V = 1000 volts (1 kV). The momentum would be

p = sqrt(2 × e × m × V)

≈ sqrt(2 × 1.6×10^-19 C × 9.11×10^-31 kg × 1000)

≈ sqrt(2.92×10^-46)

≈ 1.71×10^-23 kg·m/s.

That’s a tiny number, but in the world of particles, it’s the momentum that matters for how fast the particle travels and how it behaves when it interacts or when its wave-like nature shows up.

Relating momentum to wavelength: the wave-particle bridge

Here’s where the story gets even cooler. In quantum terms, momentum is linked to wavelength through De Broglie’s relation: λ = h/p, where h is Planck’s constant (about 6.626×10^-34 J·s). If you substitute p from our main expression, you get

λ = h / sqrt(2 m e V).

That means increasing the voltage not only shoves the particle harder; it shortens its wavelength. In practical terms, higher-speed electrons probe smaller spatial features, which is exactly why electron microscopes can resolve tiny structures with high voltages.

A quick numerical flavor for λ at 1 kV

Continuing with the electron example, using p ≈ 1.71×10^-23 kg·m/s, we get

λ ≈ h/p ≈ 6.626×10^-34 J·s / 1.71×10^-23 kg·m/s ≈ 3.9×10^-11 m,

which is about 0.04 nanometers. That’s in the range where electrons can reveal the lattice of a crystal with striking clarity. It’s also a nice illustration of how energy, momentum, and wavelength braid together in physics that you’ll encounter in both classrooms and labs.

Relativity: a quick caveat for high voltages

The non-relativistic formula p = sqrt(2 m e V) assumes the electron speeds are not relativistic. If you push electrons to very high voltages (tens of kilovolts, maybe more), their speeds creep up toward a significant fraction of the speed of light. In that regime, you’d replace the simple relation with a relativistic one; momentum and energy take a more careful dance. The beauty of the simple expression is that it nails the non-relativistic case cleanly, which is perfect for most NEET-level problems and many lab setups.

Where this pops up in the broader physics neighborhood

• Cathode ray tubes and electron beams: The same momentum–voltage link governs how fast electrons shoot through a tube and where they hit.

• Electron microscopes: As voltage climbs, wavelength shrinks; you get finer resolution. The physics is the same, just wired into a different instrument.

• Photoelectric effect and wave-particle duality: Momentum and energy concepts intertwine with how light interacts with matter. The idea that energy from a field can convert into kinetic energy—and thus momentum—parallels how photons transfer energy and momentum to electrons.

A practical mindset: how to handle questions like this on exams or in quick thinking

• Remember the core equation: eV = p^2/(2m). It’s the bridge between energy gained in an electric field and motion.

• From there, isolate p: p = sqrt(2 m e V). This is the momentum expression you’ll want most often.

• If you’re asked for a wavelength, bring in De Broglie: λ = h/p, which gives λ = h / sqrt(2 m e V).

• Check units: p is in kg·m/s; m is kg; e is C; V is V (J/C). The product m e V has units of kg·J, which matches p^2. It’s a nice consistency check.

A few mental shortcuts that keep the math quick

• p scales with the square root of V. Double the voltage, and p goes up by a factor of sqrt(2) ≈ 1.414, not by a full doubling.

• For the same voltage, heavier particles need more energy to achieve the same momentum. That’s because p^2 contains m as a factor, so more mass makes p bigger for the same energy, or equivalently, for the same p you’d need more energy to accelerate a heavier particle.

• If you know the kinetic energy, you can find momentum with p = sqrt(2 m K.E.). If you know the voltage, substitute K.E. = eV. The chain is seamless.

A small caveat about everyday intuition

It’s tempting to treat voltage as a kind of “speed booster” that directly sets velocity. In reality, momentum is what the field bestows, and velocity comes out of that momentum when you divide by the mass and account for the non-relativistic or relativistic regime. It’s a subtle distinction, but one that matters when you’re juggling multiple concepts at once—or solving a multi-step problem in a timed setting.

A compact recap for quick recall

• Core relation: eV = p^2/(2m) → p = sqrt(2 m e V)

• Wavelength link: λ = h/p = h / sqrt(2 m e V)

• Practical takeaway: p grows with the square root of the accelerating voltage; heavier particles require more energy for the same momentum; higher voltages push wavelengths shorter (in the non-relativistic domain).

• Relativity note: at very high voltages, non-relativistic formulas need adjustment; keep an eye on the speed limit.

A final thought to linger on

Physics often gives you a clean, compact answer that sits at the crossroads of energy, motion, and wave behavior. The momentum expression for a charged particle in an electric field is one of those tidy little machines. It’s not just an algebraic trick; it’s a doorway to understanding how particles move in devices you might encounter in labs, exams, or research someday. When you see a voltage, think: energy is changing hands, and momentum is the measure of how hard that change is pushing you forward. The rest follows—either to a sharper electron image in an instrument, or to a clearer wavelength that reveals the world at a scale we can’t see with the naked eye.

Key takeaways, in a line

• p = sqrt(2 e m V) is the momentum of a charged particle accelerated by voltage V.

• This p links directly to wavelength via λ = h/p.

• The relationships hold best in the non-relativistic realm; beware at very high voltages.

If you’re curious, try a quick check: plug in a proton instead of an electron, with the same 1 kV acceleration. The mass is vastly larger, so p will be bigger for the electron than for the proton under the same voltage, and the corresponding wavelength will be shorter for the electron. It’s a tiny world, but the same rules apply—and that consistency is what makes physics feel elegant, almost like a well-tuned instrument.

And that, in a nutshell, is why p = sqrt(2 e m V matters. It’s the compact bridge between energy, momentum, and the whisper of a wavelength—exactly the kind of concept that makes the physics of motion so satisfying to study.