Continuous X-ray Spectrum and Cutoff Wavelength
In an X-ray tube, electrons accelerated through voltage V reach kinetic energy eV. Energy conservation sets the maximum photon energy h nu_max = eV and hence the Duane-Hunt cutoff wavelength.
Why this shows up in the exam
Choosing X-ray tube voltage · Finding continuous-spectrum cutoff · Relating incident electron wavelength to emitted X-rays
Learn the idea
The shortest X-ray wavelength occurs when one electron gives all its kinetic energy to one photon. Most target electrons lose energy in several steps and produce a continuous spectrum. The extreme short-wavelength photon is the rare case where one electron's full qV energy becomes one photon.
🧠 Memory hook: Higher tube voltage gives a shorter cutoff wavelength.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- h nu_max = e V — maximum bremsstrahlung photon energy
- lambda_min = h c/(e V) — short-wavelength cutoff of the continuous spectrum
How to approach it
- 1Convert accelerating voltage to maximum electron energy
- 2Set only the maximum photon energy equal to eV
- 3Use lambda_min = h c/(eV) with consistent units
Common slip-ups that cost marks
- •Calling the cutoff a maximum wavelength
- •Making cutoff wavelength depend on target material
- •Equating every emitted photon energy to eV
🌟 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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