Charged Particle Accelerated Through Potential
For a non-relativistic particle accelerated from rest through a potential difference V, K = absolute q times V. Combining this with the de Broglie relation yields lambda = h divided by the square root of 2m absolute q V.
Why this shows up in the exam
Electron guns · Electron microscopes · Accelerator wavelength comparisons
Learn the idea
A charged particle accelerated from rest gains kinetic energy qV and a wavelength set by mass, charge, and voltage. A potential difference gives the particle electrical energy. That energy becomes momentum, so a larger accelerating voltage produces a larger momentum and a shorter matter wavelength.
🧠 Memory hook: Voltage raises momentum, so wavelength falls as one over root V.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- K = |q| V — kinetic-energy gain in magnitude when accelerated from rest
- lambda = h / sqrt(2 m |q| V) — non-relativistic wavelength after acceleration through V
- lambda_e(in angstrom) approximately 12.27/sqrt(V) — electron shortcut for non-relativistic accelerating voltage in volts
How to approach it
- 1Check whether the particle starts from rest
- 2Set kinetic energy gain equal to absolute q times V
- 3Use mass-charge-voltage ratios before numerical substitution
Common slip-ups that cost marks
- •Dropping the particle charge from the formula
- •Using the electron shortcut for another particle
- •Ignoring initial kinetic energy when the particle does not start from rest
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
Photons of energy 5 eV illuminate a metal of work function 2 eV. Find the stopping potential.
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