Mini Physics
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X-ray photon energy limits

Use electron energy gain eV to calculate maximum X-ray photon energy and frequency, minimum wavelength and the voltage needed for a chosen cut-off (optional H2 extension).

  • A-Level H2 Physics topic extensions

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Optional extension

These X-ray calculations are beyond the named 9478 quantum outcomes. Start with how the tube works, and recall photon energy E = hf = hc/λ and 1 eV = 1.60 × 10⁻¹⁹ J.

Start with the energy available

An electron gains kinetic energy eV through accelerating potential difference V, where e is the positive elementary-charge magnitude. We assume negligible initial kinetic energy, a steady accelerating voltage and an ideal energy limit; small recoil corrections are neglected.

A photon cannot receive more energy than the incoming electron supplies. Therefore

E_γ ≤ eV.

The limiting case gives the maximum photon energy, maximum frequency and minimum wavelength:

Eₘₐₓ = eV, fₘₐₓ = eV/h, λₘᵢₙ = hc/eV.

Most photons have less energy than this limit. Heating and multiple interactions describe typical energy sharing; they do not change the ideal energy-conservation bound into a statement about average photon energy. A spectrum can contain photons with longer wavelengths, but not shorter wavelengths than this bound in the stated model.

Choose one unit system

QuantitySI calculationElectron-volt calculation
Electron energy gaineV in joules, with V in voltsAn electron gains V eV through V volts
Example: 50 kV8.0 × 10⁻¹⁵ J50 keV
Wavelength relationλ = hc/E with E in joulesλ(nm) ≈ 1240/E(eV)

A volt is a unit of potential difference. An electron-volt is a unit of energy. Their numerical correspondence for one electron does not make their units interchangeable. Use the same energy units as the supplied value of h or hc.

Check the direction of a change

X-ray cut-off wavelength and accelerating voltage

The ideal minimum wavelength decreases inversely with accelerating voltage, from 0.0620 nm at 20 kV through 0.0248 nm at 50 kV to 0.0124 nm at 100 kV.

Scroll across the graph to read all labels.

The ideal minimum wavelength decreases inversely with accelerating voltage, from 0.0620 nm at 20 kV through 0.0248 nm at 50 kV to 0.0124 nm at 100 kV.The ideal minimum wavelength decreases inversely with accelerating voltage, from 0.0620 nm at 20 kV through 0.0248 nm at 50 kV to 0.0124 nm at 100 kV.
Calculated ideal limit with hc ≈ 1240 eV nm: λmin(nm) = 1.240/V(kV). Doubling voltage halves the shortest wavelength; this is not the wavelength of every emitted photon.
Open full-size graph
View figure data
Values for X-ray cut-off wavelength and accelerating voltage
Accelerating voltage (kV)Ideal wavelength limit
200.062
220.056363636363636366
240.051666666666666666
260.047692307692307694
280.04428571428571428
300.04133333333333333
320.03875
340.036470588235294116
360.034444444444444444
380.03263157894736842
400.031
420.029523809523809525
440.028181818181818183
460.026956521739130435
480.025833333333333333
500.0248
520.023846153846153847
540.022962962962962963
560.02214285714285714
580.021379310344827585
600.020666666666666667
620.02
640.019375
660.018787878787878787
680.018235294117647058
700.017714285714285714
720.017222222222222222
740.016756756756756756
760.01631578947368421
780.015897435897435898
800.0155
820.015121951219512195
840.014761904761904763
860.01441860465116279
880.014090909090909091
900.013777777777777778
920.013478260869565217
940.013191489361702127
960.012916666666666667
980.012653061224489795
1000.0124

At fixed other conditions, doubling voltage doubles the ideal maximum photon energy and frequency, and halves the minimum wavelength. The graph uses hc ≈ 1240 eV nm, so λₘᵢₙ(nm) = 1.240/V(kV). The inverse curve helps distinguish this wavelength relation from a direct proportion.

If you are given a cut-off wavelength and asked for voltage, first obtain the maximum photon energy from hc/λₘᵢₙ, then equate it to eV.

Worked calculations

Worked 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.
Worked solution

  1. State the ideal maximum

    Method

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

    Reason

    The ideal endpoint corresponds to one photon receiving the available electron energy, with recoil neglected.

    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.
Show 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

Spot the mistake 3

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

About 5 min

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

Show 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

Guided practice 4

Maximum frequency from accelerating voltage

About 6 min

Try the calculation

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.

Try this before viewing the solution

Hints

Hint 1: choose the energy relation
First find the electron energy in joules, then use Eₘₐₓ = hfₘₐₓ.
Show solution step by step

  1. Convert voltage

    Method

    40 kV is 4.0 × 10⁴ V.

    Reason

    Use SI units with coulombs.

    Working

    V = 4.0 × 10⁴ V

  2. Find maximum energy

    Method

    Calculate eV.

    Reason

    This is the ideal photon-energy maximum.

    Working

    Eₘₐₓ = 6.4 × 10⁻¹⁵ J

  3. Relate energy and frequency

    Method

    Use Eₘₐₓ = hfₘₐₓ.

    Reason

    Photon energy is proportional to frequency.

    Working

    fₘₐₓ = Eₘₐₓ/h

  4. Calculate

    Method

    Evaluate and state hertz.

    Reason

    Frequency has units s⁻¹.

    Working

    fₘₐₓ = 9.7 × 10¹⁸ Hz

Try it yourself

Mind stretcher 1: Apply the same limit at 100 kVExtension

A tube runs at 100 kV. Find its ideal cut-off wavelength using h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and e = 1.60 × 10⁻¹⁹ C. Is this the wavelength of every emitted photon?

Show reasoning

V = 1.00 × 10⁵ V and λₘᵢₙ = hc/(eV) ≈ 1.24 × 10⁻¹¹ m = 0.0124 nm. This is the shortest-wavelength limit; the tube emits a distribution containing longer wavelengths. Use the spectrum lesson to distinguish this endpoint from characteristic lines.

Syllabus and review details

No official syllabus alignment is listed for this lesson.