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Coolidge X-ray Tube

Key idea: Understand how a Coolidge X-ray tube produces X-rays, and how accelerating voltage sets the maximum photon energy (optional enrichment for A Level Physics).

  • A-level H2 Physics topic extensions
On this page

Learning objectives

  • Explore potential-barrier transmission, tunnelling and X-ray production and spectra beyond the H2 syllabus.
Legacy syllabus material

X-ray tube construction and minimum-wavelength calculations are not named in the current 9478 Quantum Physics outcomes. Use this page only if your school or an older paper requires the topic.

Optional / Enrichment (9478 scope)

X-ray production is not a core H2 Physics (9478) learning outcome. Use this page as enrichment to connect photon energy (E = hf) to accelerating potential (eV).

1. Definitions (Must Know)

A. X-rays

X-rays are electromagnetic waves with very short wavelength (roughly 0.01 nm to 10 nm), so individual photons have high energy.

B. Coolidge X-ray tube

A Coolidge X-ray tube produces X-rays by:

  • heating a cathode to emit electrons (thermionic emission),
  • accelerating the electrons through a large potential difference V,
  • decelerating them rapidly in a metal target (anode), converting energy into X-ray photons.

C. Accelerating potential, V

When an electron is accelerated through potential difference V, it gains kinetic energy: Δ E = eV where e = 1.60 × 10⁻¹⁹ C.

2. Key Ideas (What Earns Marks)

  • The largest possible X-ray photon energy is about the electron’s kinetic energy gain: Eₘₐₓ ≈ eV
  • So the shortest (cut-off) wavelength satisfies: λₘᵢₙ = hc/eV
  • Only a small fraction of the electron energy becomes X-rays; most becomes heat in the target.

3. Detailed Explanations

In an evacuated tube, a heated negative cathode emits electrons. A high potential difference accelerates the electron beam towards a positive angled metal target in the anode. Most electron energy heats the target and a small fraction leaves as X-rays.
A Coolidge tube converts electron kinetic energy into mostly target heating and a small X-ray output. The diagram is schematic and omits real shielding and engineering controls.

A. Why a large potential difference matters

If the accelerating voltage is larger, each electron arrives with higher kinetic energy (eV). That increases:

  • the maximum possible photon energy (so λₘᵢₙ becomes smaller),
  • the overall X-ray intensity (more energy available per electron).

B. Why the target (anode) gets hot

Rapid deceleration in the target transfers energy to the lattice (atomic vibrations), so most input electrical energy becomes thermal energy in the target.

C. Material choices (engineering, not examinable)

Targets are chosen to manage heating and to improve X-ray output.

Common design goals:

  • high melting point (to survive heating),
  • good thermal conductivity (to spread heat),
  • often high atomic number (increases X-ray production efficiency).
Safety

X-rays are ionising radiation. Production and shielding are tightly controlled in real equipment.

4. Common Mistakes

  • Mixing up electron-volt (eV) (a unit of energy) with voltage V (potential difference).
  • Thinking every electron produces a single photon of energy eV (in reality, there is a distribution of photon energies).
  • Forgetting the cut-off wavelength depends on V, not on the filament temperature.

5. Exam Tips

  • If a question gives an accelerating voltage and asks for the maximum photon energy or minimum wavelength, start with:
    • Eₘₐₓ = eV
    • λₘᵢₙ = hc/eV
  • Keep units consistent: V in volts, h in J·s, c in m/s, so λ comes out in metres.

6. Worked Examples

Modelled example 1

Minimum wavelength from accelerating voltage

Core

Problem

A Coolidge tube operates at 50 kV. Estimate its minimum X-ray wavelength using h = 6.63 × 10⁻³⁴ J s and c = 3.00 × 10⁸ m s⁻¹.
Study the worked solution

  1. State the ideal maximum

    Method

    Set maximum photon energy equal to electron kinetic-energy gain.

    Reason

    The cut-off case assumes one photon receives almost all eV.

    Working

    hc/λₘᵢₙ = eV

  2. Convert voltage

    Method

    50 kV is 5.0 × 10⁴ V.

    Reason

    SI substitution with charge in coulombs requires volts.

    Working

    V = 5.0 × 10⁴ V

  3. Calculate

    Method

    Rearrange and substitute.

    Reason

    The result is the shortest, highest-energy wavelength.

    Working

    λₘᵢₙ = hc/eV = 2.5 × 10⁻¹¹ m = 0.025 nm

Guided practice 2

Maximum photon energy

About 4 min

Problem

Find the maximum photon energy in joules for a tube operating at 30 kV.

Try this before viewing the solution

Hints

Hint 1: convert kilovolts
Use V = 3.0 × 10⁴ V in Eₘₐₓ = eV.
View solution step by step

  1. Convert

    Method

    Write kilovolts as volts.

    Reason

    The coulomb-volt product gives joules.

    Working

    30 kV = 3.0 × 10⁴ V

  2. Calculate

    Method

    Multiply by the elementary charge.

    Reason

    This is the maximum electron energy available to one photon.

    Working

    Eₘₐₓ = (1.60 × 10⁻¹⁹)(3.0 × 10⁴) = 4.8 × 10⁻¹⁵ J

Common misconception 3

Voltage needed for a target cut-off wavelength (eV·nm shortcut)

Find and correct the mistake

Learner claim

For λₘᵢₙ = 0.020 nm, a learner calculates 62 keV and reports the accelerating potential as “62 keV”. Diagnose the unit and find the required voltage using hc = 1240 eV·nm.

Try this before viewing the solution

Unit of accelerating potential

View solution step by step

  1. Find photon energy

    Method

    Use the wavelength form in eV.

    Reason

    The supplied constant pairs eV with nm.

    Working

    Eₘₐₓ = 1240/0.020 = 6.2 × 10⁴ eV = 62 keV

  2. Map energy to voltage

    Method

    A single electron gains V eV through V volts.

    Reason

    Electronvolt is an energy unit; volt is potential difference.

    Working

    62 keV ↔ 62 kV

  3. State the answer

    Method

    The required accelerating potential is approximately 62 kV.

    Reason

    That potential supplies up to 62 keV per electron.

    Working

    V ≈ 62 kV

Examiner practice 4

Maximum frequency from accelerating voltage

4 marks

Examination question

A tube operates at 40 kV. Estimate its maximum X-ray frequency using h = 6.63 × 10⁻³⁴ J s and e = 1.60 × 10⁻¹⁹ C. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Convert voltage

    1 mark

    Method

    40 kV is 4.0 × 10⁴ V.

    Reason

    Use SI units with coulombs.

    Working

    V = 4.0 × 10⁴ V

  2. Find maximum energy

    1 mark

    Method

    Calculate eV.

    Reason

    This is the ideal photon-energy maximum.

    Working

    Eₘₐₓ = 6.4 × 10⁻¹⁵ J

  3. Relate energy and frequency

    1 mark

    Method

    Use Eₘₐₓ = hfₘₐₓ.

    Reason

    Photon energy is proportional to frequency.

    Working

    fₘₐₓ = Eₘₐₓ/h

  4. Calculate

    1 mark

    Method

    Evaluate and state hertz.

    Reason

    Frequency has units s⁻¹.

    Working

    fₘₐₓ = 9.7 × 10¹⁸ Hz

Challenge 5

What changes when you increase V?

Minimal support

Independent transfer

The accelerating voltage of an X-ray tube increases while other controls are unchanged. Predict what happens to (i) λₘᵢₙ and (ii) overall X-ray intensity, with a reason for each.

Try this before viewing the solution

Hints

Hint 1: separate photon limit from output
Orient with λₘᵢₙ = hc/(eV) for the cut-off and electron energy per arrival for the qualitative output.
View solution step by step

  1. Cut-off

    Method

    λₘᵢₙ decreases.

    Reason

    A larger V gives a larger maximum photon energy, and wavelength is inversely related.

    Working

    λₘᵢₙ ∝ 1/V

  2. Intensity

    Method

    Overall X-ray intensity typically increases.

    Reason

    Each electron arrives with more energy, increasing available X-ray output under the stated unchanged controls.

    Working

    eV↑ ⇒ available energy per electron↑

7. Mind Stretchers

Mind stretcher 1: What changes if you increase filament current?Extension

Show Answer

Increasing filament current increases the number of electrons emitted per second, so X-ray intensity increases. The cut-off wavelength λₘᵢₙ is set mainly by accelerating voltage V, so it stays (approximately) unchanged.

Mind stretcher 2: Why does the target heat up so much?Extension

Explain why most of the electrical energy ends up as heat in the target rather than X-rays.

Show Answer

Most incoming electrons undergo many inelastic collisions with the target material, transferring energy to the lattice (atomic vibrations), which becomes thermal energy.

Only a small fraction of interactions produce X-ray photons, so X-ray production is inefficient and heating dominates.

8. Optional (Enrichment)

A. Why vacuum matters

The tube is evacuated so electrons can accelerate without frequent collisions with gas molecules (which would reduce electron energy and waste power as heat/light).

Continue with the next resource in this course.

Course and syllabus information
Course
A-level H2 Physics topic extensions
Syllabus scope
Beyond the syllabus
Edition
A-level H2 Physics topic extensions