Explain X-ray production and calculate the minimum wavelength
Describe electron bombardment of a metal target and use energy conservation to find the shortest X-ray wavelength.
Section 24 · Medical physics
A Level · Cambridge International Physics 9702, 2025–2027 syllabus. Topic 24 was checked against Cambridge’s 2028–2030 syllabus; the other remaining syllabus-year changes are still being reconciled. Online study supports the theory but cannot replace supervised practical work, full timed assessment or human subject-teacher review.
Before you begin
- Use exponential functions, wave speed, energy conservation and SI units; convert between joules and electronvolts when required.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What happens to the electrons before they strike the X-ray tube target?
What you will learn
- Explain how a metal target produces X-rays
- Relate accelerating p.d. to electron energy
- Calculate the minimum X-ray wavelength using hc = eVλmin
Accelerated electrons transfer energy at the target
A large potential difference accelerates electrons towards a metal target. When they hit the target, they are rapidly decelerated and can emit X-ray photons. The target also heats, so energy is not all converted to X-rays.
An electron accelerated through potential difference V gains kinetic energy eV, where e is the elementary charge. The maximum energy of one emitted photon occurs in the limiting case where one electron transfers all that kinetic energy to that photon.
The energy limit gives a minimum wavelength
A photon has energy E = hf = hc/λ. Setting its maximum energy equal to eV gives eV = hc/λmin, so λmin = hc/(eV). A larger accelerating p.d. gives a shorter minimum wavelength.
Use joules in eV and h, c in SI units to obtain λ in metres. This is a minimum wavelength; the emitted X-ray spectrum also contains photons with lower energies and longer wavelengths.
Pause and explain
Which expression gives the minimum X-ray wavelength?
Worked example
Find λmin for electrons accelerated through 50 kV. Use h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and e = 1.60 × 10⁻¹⁹ C.
Show the worked solution
- Convert 50 kV to V = 5.0 × 10⁴ V.
- λmin = hc/(eV) = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(1.60 × 10⁻¹⁹ × 5.0 × 10⁴).
- λmin = 2.49 × 10⁻¹¹ m, or 0.0249 nm.
Answer 2.49 × 10⁻¹¹ m (about 0.0249 nm).
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Electrons are accelerated through 40 kV. Calculate the maximum kinetic energy gained by one electron in joules, using e = 1.60 × 10⁻¹⁹ C.
Give me a hint
Use ΔE = eV and convert kilovolts to volts.
Compare my reasoning
- V = 40 000 V.
- ΔE = (1.60 × 10⁻¹⁹)(4.0 × 10⁴) J.
- ΔE = 6.4 × 10⁻¹⁵ J.
6.4 × 10⁻¹⁵ J.
Look for these in your work
- Converts kV to V
- Uses eV
- Reports joules
Calculate λmin at 40 kV using hc = 1.99 × 10⁻²⁵ J m and e = 1.60 × 10⁻¹⁹ C.
Give me a hint
Substitute into hc/(eV).
Compare my reasoning
- V = 4.0 × 10⁴ V.
- eV = 1.60 × 10⁻¹⁹ × 4.0 × 10⁴ = 6.4 × 10⁻¹⁵ J.
- λmin = 1.99 × 10⁻²⁵ / 6.4 × 10⁻¹⁵ = 3.11 × 10⁻¹¹ m.
3.11 × 10⁻¹¹ m (0.0311 nm).
Look for these in your work
- Finds eV in joules
- Uses λmin = hc/eV
- Gives a sensible wavelength unit
An X-ray tube’s accelerating p.d. is increased from 40 kV to 80 kV. What happens to the minimum wavelength?
Give me a hint
Use the inverse proportionality between λmin and V.
Compare my reasoning
- λmin = hc/(eV), so with h, c and e fixed, λmin ∝ 1/V.
- Doubling V doubles the maximum photon energy.
- The new minimum wavelength is half the old value.
It halves, because λmin is inversely proportional to accelerating p.d.
Look for these in your work
- Identifies inverse proportionality
- Recognises the p.d. doubles
- States the resulting wavelength change
Common mistakes
- Using 50 instead of 50 000 volts
- Using λ = hc/eV without converting the accelerating p.d. to volts
- Calling λmin the wavelength of every X-ray photon
What can you explain now?
What happens to the electrons before they strike the X-ray tube target?
Compare with the explanation
They are accelerated by a potential difference
The electrical energy gain is eV.
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Make it stick
Why does a larger accelerating p.d. allow a smaller minimum X-ray wavelength?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.