X-rays — the Continuous Spectrum, Characteristic Lines and Moseley's Law

Fire fast electrons into a metal and two quite different things happen, and the X-ray spectrum that comes out shows both. Most electrons are slowed gradually by the nuclei they pass, and each deceleration throws off a photon of any energy up to the electron's whole kinetic energy — a smooth hump with a sharp short-wavelength edge. A few knock an electron out of an inner shell of a target atom, and the vacancy is filled by an outer electron that emits a photon of one exact energy — sharp lines that identify the element. The edge is set by the accelerating voltage alone; the lines are set by the target alone. Keeping those two facts apart answers most questions on the topic, and Moseley's law turns the second into a way of counting protons.

Producing X-rays and the Continuous Spectrum

In a Coolidge tube a heated filament emits electrons, a potential difference V of tens of kilovolts accelerates them across an evacuated tube, and they strike a heavy-metal target. Each electron arrives with kinetic energy eV. Deceleration in the field of the target nuclei (bremsstrahlung, "braking radiation") produces photons whose energy can be anything up to eV, so the spectrum is continuous. The most energetic photon possible takes all of it, which fixes the cut-off wavelength: λ_min = hc/(eV), or with hc = 1240 eV nm, λ_min (nm) = 1240/V (volts).

🎯 Which knob changes what
The accelerating voltage sets the electrons' energy, so it sets λ_min (λ_min ∝ 1/V) and the hardness of the beam. The filament current sets how many electrons arrive per second, so it changes the intensity of the whole spectrum and nothing about its wavelengths. The target material sets the characteristic lines. A question that changes one of these and offers an effect belonging to another is the standard trap.

Characteristic Lines

If an incoming electron ejects a K-shell (n = 1) electron of a target atom, an electron from a higher shell drops into the vacancy and emits a photon equal to the energy difference. L → K gives the Kα line, M → K the Kβ line; vacancies in the L shell give the L series. Because M is farther from K than L is, Kβ has more energy and a shorter wavelength than Kα. These lines appear only once eV exceeds the K-shell binding energy, but once they appear their wavelengths do not move when V is raised further — they belong to the atom.

Continuous versus characteristic X-rays
FeatureContinuous spectrumCharacteristic lines
OriginDeceleration of electrons near nucleiElectron transitions into inner-shell vacancies
Set byAccelerating voltage (λ_min = hc/eV)Atomic number of the target
Raise Vλ_min shifts to shorter wavelengthWavelengths unchanged (lines may brighten)
Raise filament currentMore intense, same λ_minMore intense, same wavelengths

Moseley's Law — Counting Protons with X-rays

Moseley measured the Kα frequency for element after element and found √ν = a(Z − b): the square root of the frequency is a straight-line function of the atomic number Z. For Kα the screening constant is b ≈ 1 — the one remaining K electron shields one unit of nuclear charge from the L electron that falls in — and a Bohr-type estimate gives ν(Kα) ≈ (3/4) R c (Z − 1)², or an energy of about 13.6 × (3/4)(Z − 1)² eV. For copper, Z = 29, that is 13.6 × 0.75 × 784 ≈ 8.0 keV, close to the measured value.

📖 Why this mattered
Before Moseley, elements were ordered by atomic mass, and a few pairs (cobalt and nickel, argon and potassium) sat awkwardly. His straight line showed that the true ordering quantity is the nuclear charge Z, and gaps in the line predicted elements not yet found. The modern periodic law — properties are periodic functions of atomic number — rests on this measurement.
⚠️ Scaling between elements
Kα energy scales as (Z − 1)², not Z², so its wavelength scales as 1/(Z − 1)². For Z = 11 and Z = 21 the wavelength ratio is (20/10)² = 4, while using Z² would give (21/11)² ≈ 3.6. The "− 1" matters most for light elements.

Key takeaways

  • The continuous spectrum comes from braking electrons and ends sharply at λ_min = hc/(eV) = 1240/V nm, set by the accelerating voltage alone.
  • Characteristic lines come from inner-shell vacancies and are fixed by the target; raising V does not move them.
  • Kβ (M → K) has more energy and a shorter wavelength than Kα (L → K).
  • Filament current changes intensity only; voltage changes λ_min; target changes the lines.
  • Moseley: √ν = a(Z − b), with b ≈ 1 for Kα, so E(Kα) ≈ 10.2 (Z − 1)² eV and wavelengths scale as 1/(Z − 1)².

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. The cut-off (minimum) wavelength of the continuous X-ray spectrum from a Coolidge tube depends on:

    1. The target material only
    2. The accelerating voltage only
    3. The filament current only
    4. Both the target material and the filament current
    Show answer

    Answer: B — The accelerating voltage only

    The shortest-wavelength photon takes an electron's entire kinetic energy eV, so λ_min = hc/(eV) depends only on the accelerating voltage. The target decides the characteristic lines, and the filament current decides how many electrons arrive, hence the intensity — neither moves the cut-off.

  2. Electrons accelerated through 20 kV strike a target. Taking hc = 1240 eV nm, the cut-off wavelength of the X-rays produced is:

    1. 62 pm
    2. 6.2 pm
    3. 620 pm
    4. 62 nm
    Show answer

    Answer: A — 62 pm

    λ_min = 1240/20 000 nm = 0.062 nm = 62 pm. The other options are the same digits with the power of ten slipped — 62 nm would be ultraviolet, far too long for 20 keV photons, which is a useful sense-check.

  3. The accelerating voltage of an X-ray tube is doubled while the target is unchanged. Then:

    1. λ_min doubles and the Kα wavelength halves
    2. λ_min and the Kα wavelength both halve
    3. λ_min halves and the Kα wavelength is unchanged
    4. Neither changes; only the intensity rises
    Show answer

    Answer: C — λ_min halves and the Kα wavelength is unchanged

    λ_min ∝ 1/V, so doubling V halves it. The Kα line is an energy difference inside the target atom and does not depend on how hard the atom was struck, so its wavelength stays put. Halving both treats the characteristic line as if it were the spectrum's edge.

  4. According to Moseley's law, a straight line is obtained when one plots, against atomic number Z:

    1. The frequency ν of the Kα line
    2. The wavelength λ of the Kα line
    3. The square root of the Kα frequency, √ν
    4. The logarithm of the Kα frequency
    Show answer

    Answer: C — The square root of the Kα frequency, √ν

    Moseley's law is √ν = a(Z − b), linear in Z for the square root of the frequency. ν itself grows as (Z − 1)², a parabola, and λ falls as 1/(Z − 1)². Plotting ν against Z is the usual slip.

  5. Using Moseley's law with screening constant 1 for the Kα line, the ratio of the Kα wavelength of an element with Z = 11 to that of an element with Z = 21 is:

    1. 4
    2. 2
    3. 1/4
    4. (21/11)²
    Show answer

    Answer: A — 4

    λ ∝ 1/(Z − 1)², so λ(11)/λ(21) = (21 − 1)²/(11 − 1)² = 400/100 = 4: the lighter element has the longer wavelength. 1/4 inverts the ratio, 2 forgets the square, and (21/11)² drops the screening constant.

  6. Estimating with E(Kα) ≈ 13.6 × (3/4)(Z − 1)² eV, the Kα photon energy of copper (Z = 29) is about:

    1. 10.7 keV
    2. 8.6 keV
    3. 0.8 keV
    4. 8.0 keV
    Show answer

    Answer: D — 8.0 keV

    13.6 × 0.75 × 28² = 13.6 × 0.75 × 784 ≈ 7997 eV ≈ 8.0 keV. Using Z² = 841 gives about 8.6 keV, and dropping the 3/4 (treating the transition as from infinity to n = 1) gives about 10.7 keV.

  7. For the same target, how does the Kβ line compare with the Kα line?

    1. Kβ has higher photon energy and shorter wavelength
    2. Kβ has lower photon energy and longer wavelength
    3. They have equal energy but different intensity
    4. Kβ arises from an L → K transition
    Show answer

    Answer: A — Kβ has higher photon energy and shorter wavelength

    Kβ is an M → K transition and Kα an L → K one. The M shell lies higher than L, so the drop to K is larger: Kβ carries more energy and a shorter wavelength. L → K is Kα, so calling it the source of Kβ mixes up the two lines; Greek-letter order follows increasing energy within a series.

  8. The filament current of an X-ray tube is increased while the accelerating voltage is kept fixed. The effect on the X-ray spectrum is:

    1. λ_min decreases, while the characteristic lines stay where they were
    2. The characteristic lines shift to shorter wavelength, with λ_min unchanged
    3. The intensity increases at every wavelength, with λ_min and the lines unchanged
    4. The continuous part vanishes, leaving a purely characteristic spectrum
    Show answer

    Answer: C — The intensity increases at every wavelength, with λ_min and the lines unchanged

    More filament current means more electrons per second, each with the same energy eV. So more photons are produced at every wavelength, but λ_min = hc/(eV) and the characteristic wavelengths are unchanged. Expecting λ_min to fall confuses the number of electrons with their energy, the same distinction as intensity versus frequency in the photoelectric effect.