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X-Rays · Lesson 3 of 5

Characteristic X-Rays

14 min read3 ideas5 checksBy Asheshanand Acharya
What you will learn 0 / 3
Inner shell vacancies

Sharp peaks on a smooth hump

The spectrum of a molybdenum tube at \(30\,\text{kV}\) has two sharp peaks on its smooth hump. One is at \(0.063\,\text{nm}\) and a taller one at \(0.071\,\text{nm}\). Braking cannot explain them, because braking favours no particular energy. Something inside the molybdenum atom picks out these two energies. They are called characteristic X-rays, because they are characteristic of the target.

The electrons of an atom are arranged in shells. The shell with \(n = 1\) is called the K shell, \(n = 2\) the L shell and \(n = 3\) the M shell. In a heavy atom every inner shell is full. The K shell holds two electrons, deep down near the nucleus.

How deep the K shell is

In hydrogen, the electron on \(n = 1\) is bound by \(13.6\,\text{eV}\). In a heavy atom a K electron is bound far more tightly. It lies inside all the other electrons, so it feels nearly the whole nuclear charge \(Ze\). The Bohr energy grows as \(Z^2\). For molybdenum, \(Z = 42\), a rough estimate is \(13.6 \times 41^2 \approx 22.9\,\text{keV}\).

The measured value for molybdenum is \(20.0\,\text{keV}\), so the rough estimate is close. This binding energy of a K electron is called the K edge. The L and M shells are held much less tightly, \(2.5\) and \(0.2\,\text{keV}\) in molybdenum. On an energy ladder, the K shell sits far below the other two.

An electron from the filament can knock a K electron clean out of the atom. For this, it must bring at least the K edge, \(20.0\,\text{keV}\) for molybdenum. The atom is left with a vacancy in its K shell. A vacancy so deep does not last. Within a tiny fraction of a second, an electron from a higher shell falls into it.

principle 4.2
Energy of a characteristic photon

When an electron falls from a higher shell into a vacancy in a lower shell, the atom gives out one photon. Its energy is the difference between the energies of the two shells: \(h\nu = E_{\text{upper}} - E_{\text{lower}}\).

K alpha k beta and l alpha lines

Three lines and their names

The lines are named by the shell that holds the vacancy. A fall into the K shell gives a K line, and a fall into the L shell an L line. The Greek letter says where the electron came from. The letter α means from the next shell up, and β from two shells up.

Predict firstLocked

A molybdenum atom has a vacancy in its K shell. It can be filled by an L electron, giving the Kα line, or by an M electron, giving the Kβ line. Which photon has the shorter wavelength?

A vacancy in an inner shell, and the photon that fills itThe inner shells of a molybdenum atom drawn to scale as an energy ladder: the K shell at minus 20.0 keV, far below the L shell at minus 2.5 keV and the M shell at minus 0.2 keV. A fast electron from the filament, grey, flies in and knocks one electron clean out of the atom, leaving a hole. An electron from a higher shell falls into the hole, and a photon leaves carrying the difference in energy. Chip K-alpha: an L electron fills a K hole, 17.5 keV, 0.071 nanometre. Chip K-beta: an M electron fills a K hole, 19.8 keV, 0.063 nanometre. Chip L-alpha: an M electron fills an L hole, 2.3 keV, 0.54 nanometre. The other electrons stay in their places throughout.n = 1 (K)n = 2 (L)n = 3 (M)−20.0 keV−2.5 keV−0.2 keV

Play each chip of the figure above. For Kα an L electron falls into the K vacancy. The photon carries \(20.0 - 2.5 = 17.5\,\text{keV}\), and its wavelength is \(1240/17500 = 0.071\,\text{nm}\). For Kβ an M electron falls in. The photon carries \(20.0 - 0.2 = 19.8\,\text{keV}\), and \(\lambda = 0.063\,\text{nm}\).

The Lα chip starts with a vacancy in the L shell. An M electron falls into it, and the photon carries only \(2.5 - 0.2 = 2.3\,\text{keV}\). Its wavelength is \(0.54\,\text{nm}\), almost eight times that of Kα. L lines are always far softer than K lines, because the L and M shells lie close together near the top of the ladder.

RememberEnergies of the characteristic lines
\[ E_{K\alpha} = |E_K| - |E_L|, \quad E_{K\beta} = |E_K| - |E_M|, \quad E_{L\alpha} = |E_L| - |E_M|, \quad \lambda = \dfrac{hc}{E} \]
Holds whena fall of one electron into a single vacancy of a target atom; \(|E_K|, |E_L|, |E_M|\) are the binding energies of the K, L and M shells.
Symbols\(E\) the photon energy of the line, \(\lambda\) its wavelength, \(|E_K| > |E_L| > |E_M|\) the binding energies
Unitsenergies in keV; \(\lambda\) in nm with \(hc = 1.24\,\text{keV nm}\)
ExampleThree lines from one ladder

The K, L and M shells of a target atom have energies \(-20.0\), \(-2.6\) and \(-0.5\,\text{keV}\). Find the energy and wavelength of its Kα, Kβ and Lα photons.

Reveal one step at a time and finish the rest yourself.
Checkpoint · numericUnanswered

In copper the K shell is bound by \(8.98\,\text{keV}\) and the L shell by \(0.93\,\text{keV}\). Find the wavelength of copper's Kα line, in nm.

nm± 2 %
CheckpointUnanswered

Why is the Lα line of an element so much longer in wavelength than its Kα line?

Intuition

Think of the inner shells as a deep well with a short ladder at the top. The K shell is at the bottom of the well. The L and M shells are two rungs close together near the top. Any fall to the bottom is long, so the K lines are hard and close together. A step between the top rungs is short, so the L lines are soft.

The Kα peak is always taller than the Kβ peak. After a K vacancy forms, an L electron is far more likely to fill it than an M electron. In molybdenum, Kβ is only about a fifth as strong as Kα.

Peaks on the continuum

When the peaks appear

The peaks need K vacancies, and a K vacancy needs an incoming electron with at least the K edge. So the K lines of molybdenum appear only when \(eV\) exceeds \(20.0\,\text{keV}\), that is, above \(20.0\,\text{kV}\). They all appear together. At \(18\,\text{kV}\) the tube gives no Kα, even though a Kα photon carries only \(17.5\,\text{keV}\).

The X-ray spectrum as the tube voltage changesA graph of X-ray intensity against wavelength, from 0 to 0.16 nanometre. A slider sets the tube voltage from 10 to 50 kilovolts; chips choose a molybdenum or a tungsten target. The continuous spectrum is zero below a sharp cut-off wavelength, marked by a dashed line, then rises to a broad hump and falls slowly. As the voltage rises the cut-off slides to shorter wavelengths, 0.041 nanometre at 30 kilovolts and 0.025 at 50, the same for both targets, and the whole curve grows; the heights are schematic. Tungsten gives a larger continuum than molybdenum. For molybdenum two sharp peaks, K-beta at 0.063 and K-alpha at 0.071 nanometre, appear only above 20 kilovolts and never move. For tungsten, whose K lines need 69.5 kilovolts, two L lines at 0.128 and 0.148 nanometre appear above about 11 kilovolts. Readouts give the voltage, the cut-off and the lines.0.020.040.060.080.10.120.140.16Oλ (nm)intensity λmin Kα Kβ V = 30 kV, λmin = hc/eV = 0.041 nm Mo lines: Kβ 0.063 nm, Kα 0.071 nm
SimThe same spectrum, now for the peaks. With the molybdenum target, start at \(15\,\text{kV}\) and raise the voltage. Below \(20\,\text{kV}\) there are no peaks. Above it, Kβ at \(0.063\,\text{nm}\) and Kα at \(0.071\,\text{nm}\) appear and grow, but never move. Tungsten's K edge is \(69.5\,\text{keV}\), beyond this slider, so its tube shows only two softer L lines, at \(0.128\) and \(0.148\,\text{nm}\).

Once the peaks appear, a higher voltage makes them taller, as the figure above shows. More electrons now have enough energy to make K vacancies. But the peaks never move. Their wavelengths are fixed by the shell energies of the target atom, and the tube voltage cannot change those. Only a different target moves them.

Property
Continuous spectrum
Characteristic lines
Cause
braking of electrons by nuclei
falls into inner-shell vacancies
Shape
a smooth band of all wavelengths above \(\lambda_{min}\)
sharp peaks at a few wavelengths
When the voltage rises
\(\lambda_{min}\) moves to shorter wavelengths
peaks grow but stay in place
When the target changes
\(\lambda_{min}\) stays the same
peaks move to new wavelengths
Needs a least voltage?
no
yes, the K edge for the K lines
ExampleA molybdenum tube at 40 kV and 15 kV

A molybdenum tube (K edge \(20.0\,\text{keV}\)) runs first at \(40\,\text{kV}\), then at \(15\,\text{kV}\). Find the cut-off wavelength at each voltage. At which voltage does the Kα line at \(0.071\,\text{nm}\) appear?

Try it on paper first. The solution is one tap away when you are ready.
CheckpointUnanswered

A molybdenum tube (K edge \(20.0\,\text{keV}\), Kα \(17.5\,\text{keV}\)) runs at \(18\,\text{kV}\). Does its spectrum show the Kα line?

Checkpoint · select all that applyUnanswered

The molybdenum target of a tube is replaced by copper. The voltage and current stay the same. Which statements are true?

Deeper

The K edge shows up in absorption too. A metal foil absorbs X-rays weakly just below its K edge, and strongly just above it, where the photons can knock out K electrons.

In the world
Soft X-rays for a breast examination

Mammography, the X-ray examination of the breast, needs soft X-rays that tell one soft tissue from another. Many mammography tubes have a molybdenum target run at about \(28\,\text{kV}\). Most of their useful X-rays are the Kα and Kβ lines of molybdenum, near \(17.5\) and \(19.8\,\text{keV}\). A thin molybdenum filter removes much of the rest, so the beam is nearly one colour of X-ray.

Story · Charles Barkla, London, 1911
The K and L radiations

From 1906 Charles Barkla sent X-rays onto different elements and studied the X-rays they gave out in turn. Each element sent out its own kind, one harder group and one softer group. In 1911 he called them the K and L radiations. He started in the middle of the alphabet, in case harder or softer groups turned up later. None harder than K has ever been found. Barkla received the Nobel Prize in Physics in 1917.

Exam traps
  1. Thinking Kα has the shorter wavelength because α comes first.
    α is the smaller fall, L to K. The larger fall, M to K, is Kβ, so Kβ is shorter.
  2. Taking the Kα energy as the least voltage for the K lines.
    The line needs a K vacancy first, which needs the K edge. For molybdenum that is \(20.0\,\text{kV}\), not \(17.5\,\text{kV}\).
  3. Expecting the peaks to move when the voltage changes.
    The peaks are fixed by the target's shells. The voltage makes them taller or makes them vanish, never moves them.
  4. Using negative shell energies with the wrong sign.
    The photon energy is always positive: the binding energy of the lower shell minus that of the upper shell.