X-Rays and Laser Radiations

Lessons

Senior 6 · Physics

X-Rays and Laser Radiations

4 lessons

Lesson 2 of 4

The X-Ray Spectrum

25 min

A smooth hump with two sharp spikes

Point the X-rays from a tube at a crystal and turn the crystal slowly, measuring the intensity at each angle. Bragg's law (Lesson 1) turns each angle into a wavelength, so the result is a spectrum: intensity against wavelength. For a molybdenum target at 35 kV, the spectrum has a surprising shape. Nothing at all arrives below 0.035 nm; then comes a broad, smooth hump; and on top of the hump stand two very sharp, tall spikes at 0.063 nm and 0.071 nm. Two different processes are at work. This lesson explains each one, and the simple rule, found by William Duane and Franklin Hunt in 1915, that fixes the sharp edge.

Figure 2.1: The X-ray spectrum of a molybdenum target at two tube voltages (shape schematic).

The continuous spectrum: braking radiation

When a fast electron passes close to a nucleus in the target, the strong electric attraction deflects and slows it. An accelerating charge radiates, so the electron emits an X-ray photon carrying away the kinetic energy it lost. This is called bremsstrahlung, German for "braking radiation". An electron can lose any fraction of its energy in one encounter, or share it over many, so the photons have every energy up to a maximum: the spectrum is continuous.

The Duane–Hunt limit

The most energetic photon possible is made when an electron loses ALL its kinetic energy $e V$ in a single encounter. Then $hf_{\max} = eV$ , and since $f = \frac{c}{\lambda}$ ,

\frac{hc}{\lambda_{\min}} = eV \quad\Rightarrow\quad \lambda_{\min} = \frac{hc}{eV}

This is the Duane–Hunt law. The minimum wavelength depends ONLY on the accelerating voltage, not on the target metal or the tube current. With $hc = 1.24 \times 10^{-6}\,\mathrm{eV\,m}$ , a handy form is $\lambda_{\min}\,(\text{in nm}) = \frac{1.24}{V\,(\text{in kV})}$ . The existence of this sharp edge is strong evidence that X-rays come in photons: a classical wave could carry any small amount of energy at any wavelength.

Definition

Bremsstrahlung (braking radiation) is the continuous X-ray spectrum emitted when electrons are decelerated by the nuclei of the target. Its Duane–Hunt limit is the shortest wavelength, $\lambda_{\min} = \frac{hc}{eV}$ , produced when an electron gives all its kinetic energy $e V$ to one photon. Characteristic X-rays are sharp lines emitted when electrons fall into vacancies in the inner shells of the target atoms. Example: at 50 kV, $\lambda_{\min} = \frac{1.24}{50} = 0.0248\,\mathrm{nm}$ .

Worked example 1: the cut-off wavelength

(a) Find $\lambda_{\min}$ for a tube at 35 kV. (b) A spectrum has $\lambda_{\min} = 0.0248$ nm. What was the tube voltage?

Solution

(a) $\lambda_{\min} = \frac{hc}{eV} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{(1.60 \times 10^{-19})(35 \times 10^{3})} = 3.55 \times 10^{-11}\,\mathrm{m} = 0.0355\,\mathrm{nm}$ , matching Figure 2.1.

(b) $V = \frac{hc}{e\lambda_{\min}} = \frac{1.24 \times 10^{-6}\,\mathrm{V\,m}}{0.0248 \times 10^{-9}\,\mathrm{m}} = 5.0 \times 10^{4}\,\mathrm{V} = 50\,\mathrm{kV}$ .

Worked example 2: measuring Planck's constant

Duane and Hunt used the cut-off to measure $h$ . In an experiment, a tube at 40.0 kV gave $\lambda_{\min} = 0.0310$ nm. Find $h$ ( $e = 1.602 \times 10^{-19}$ C, $c = 2.998 \times 10^{8}\,\mathrm{m\,s^{-1}}$ ).

Solution

h = \frac{eV\lambda_{\min}}{c} = \frac{(1.602 \times 10^{-19})(40.0 \times 10^{3})(0.0310 \times 10^{-9})}{2.998 \times 10^{8}} = 6.63 \times 10^{-34}\,\mathrm{J\,s}

This agrees with the value from the photoelectric effect (Senior 5): the same constant governs light going OUT of a metal and X-rays coming IN.

The characteristic lines

A fast electron can also knock an electron out of an INNER shell of a target atom, leaving a vacancy in the K shell. An electron from a higher shell immediately drops into the gap, and the energy difference leaves as a single X-ray photon:

L shell to K shell: the $\mathrm{K}_\alpha$ line;
M shell to K shell: the $\mathrm{K}_\beta$ line (higher energy, shorter wavelength).

Figure 2.2: Inner energy levels of molybdenum (rounded) and the transitions that give its K lines.

The energy levels belong to the target element, so the lines appear at the same wavelengths whatever the tube voltage: they are the element's X-ray fingerprint. They can only appear if each electron has enough energy to knock out a K electron: $e V$ must exceed the K-shell binding energy (20.0 keV for molybdenum, 69.5 keV for tungsten).

Moseley's law

In 1913 Henry Moseley measured the $\mathrm{K}_\alpha$ lines of many elements and found that $\sqrt{f}$ rises in equal steps with atomic number. A simple Bohr-model argument explains it: a K-shell vacancy leaves the L electron "seeing" a nuclear charge of about $(Z - 1) e$ (one K electron still screens the nucleus), and the jump from $n = 2$ to $n = 1$ then has energy

E_{K\alpha} \approx 13.6\,\mathrm{eV} \times (Z-1)^2\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 10.2\,\mathrm{eV} \times (Z-1)^2

Moseley's work showed that the order of the periodic table is set by $Z$ , the charge of the nucleus, not by atomic mass, and it revealed gaps for elements not yet discovered.

Worked example 3: lines from energy levels and from Moseley's law

(a) Using Figure 2.2, find the energy and wavelength of the molybdenum $\mathrm{K}_\alpha$ line. (b) Use Moseley's law to estimate the $\mathrm{K}_\alpha$ energy and wavelength for copper ( $Z = 29$ ).

Solution

(a) $E = 20.0 - 2.5 = 17.5\,\mathrm{keV}$ ; $\lambda = \frac{1.24 \times 10^{-6}}{17.5 \times 10^{3}} = 7.09 \times 10^{-11}\,\mathrm{m} = 0.0709\,\mathrm{nm}$ , the tall line in Figure 2.1.

(b) $E \approx 10.2 \times (29 - 1)^2 = 10.2 \times 784 = 8.0 \times 10^{3}\,\mathrm{eV} = 8.0\,\mathrm{keV}$ ; $\lambda = \frac{1.24 \times 10^{-6}}{8.0 \times 10^{3}} = 0.155\,\mathrm{nm}$ . The measured copper $\mathrm{K}_\alpha$ wavelength is 0.154 nm, the X-ray wavelength most used in crystallography laboratories.

Changing the tube settings

Change	$\lambda_{\min}$	Continuous spectrum	Characteristic lines
raise the voltage	shorter	more intense, extends to shorter λ	same wavelengths, stronger
raise the tube current	unchanged	more intense at every λ, same shape	same wavelengths, stronger
use a target of higher Z	unchanged	more intense	new lines, at shorter wavelengths

How the spectrum is measured

A narrow beam from the tube falls on a crystal of known plane spacing $d$ on a turntable. A detector (an ionisation chamber or GM tube) moves round at twice the crystal's angle, so it always catches the reflected beam. At each angle $\theta$ , the intensity is recorded and $\lambda = \frac{2d\sin\theta}{n}$ is calculated. This is a university or industrial instrument (an X-ray spectrometer or diffractometer), used in shielded rooms.

Warning

Common mistakes. (1) Thinking $\lambda_{\min}$ depends on the target: it depends only on $V$ . (2) Thinking the characteristic lines move when the voltage changes: their wavelengths are fixed by the target atoms. (3) Writing $\lambda_{\min} = \frac{hc}{V}$ : the energy is $e V$ , so $\lambda_{\min} = \frac{hc}{eV}$ . (4) Confusing these lines with the visible line spectra of Senior 5: they come from INNER shells, with energies of keV, not eV.

Tip

The edge belongs to the voltage; the spikes belong to the metal. And for quick numbers: $\lambda_{\min}$ in nm $= \frac{1.24}{V}$ with $V$ in kV, and photon energy in keV $= \frac{1.24}{\lambda}$ with $\lambda$ in nm.

X-ray production animation (A Level Physics HQ). Watch how braking radiation gives the continuous spectrum and how inner-shell transitions give the characteristic lines.

✦

Practice

Explain why the X-ray spectrum has a sharp minimum wavelength.
Find $\lambda_{\min}$ for tube voltages of 25 kV and 100 kV.
An X-ray photon has energy 59.3 keV. Find its wavelength.
A tungsten-target tube is run at 50 kV. Will tungsten's K lines (which need at least 69.5 keV) appear? Explain.
Sketch the spectrum of Figure 2.1 and, on the same axes, the spectrum when the tube current is doubled at the same voltage.
Use Moseley's law to estimate the $\mathrm{K}_\alpha$ energy of iron ( $Z = 26$ ). The measured value is 6.40 keV; find the percentage difference.
The $\mathrm{K}_\beta$ line of a metal is at 0.139 nm and its $\mathrm{K}_\alpha$ line at 0.154 nm. Find the energy difference between its L and M shells.
Mutesi says: "If we use a target made of a heavier metal, the shortest X-ray wavelength will become shorter." Evaluate her statement.

Show answersHide answers

A photon cannot have more energy than one electron's kinetic energy $e V$ . The largest photon energy, given when an electron loses all its energy at once, corresponds to $\lambda_{\min} = \frac{hc}{eV}$ ; no shorter wavelength is possible.
25 kV: $\frac{1.24}{25} = 0.0496\,\mathrm{nm}$ ; 100 kV: 0.0124 nm.
$\lambda = \frac{1.24 \times 10^{-6}}{59.3 \times 10^{3}} = 2.09 \times 10^{-11}\,\mathrm{m} = 0.0209\,\mathrm{nm}$ .
No. Each electron has at most 50 keV, less than the 69.5 keV needed to remove a K electron from tungsten, so there are no K-shell vacancies and no K lines (L lines can still appear).
The new curve has the same $\lambda_{\min}$ and lines at the same wavelengths, but every intensity is doubled.
$10.2 \times 25^2 = 6.38\,\mathrm{keV}$ ; difference $\frac{6.40 - 6.38}{6.40} = 0.3\%$ .
$E_{K\beta} = \frac{1.24}{0.139} = 8.92\,\mathrm{keV}$ ; $E_{K\alpha} = \frac{1.24}{0.154} = 8.05\,\mathrm{keV}$ . L to M difference $= 8.92 - 8.05 = 0.87\,\mathrm{keV}$ .
Wrong. $\lambda_{\min} = \frac{hc}{eV}$ depends only on the accelerating voltage. A heavier target gives more intense bremsstrahlung and different characteristic lines, but the same cut-off.

Key points

Continuous spectrum: bremsstrahlung from electrons decelerated by nuclei.
Duane–Hunt limit: $\lambda_{\min} = \frac{hc}{eV}$ , set only by the voltage; evidence for photons.
Characteristic lines ( $\mathrm{K}_\alpha$ : L to K; $\mathrm{K}_\beta$ : M to K) are fixed by the target element; they need $e V$ above the K-shell binding energy.
Moseley: $E_{K\alpha} \approx 10.2\,\mathrm{eV} \times (Z-1)^2$ .

Sources and credits

30.4: X Rays: Atomic Origins and Applications, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Video: X Ray Production Animation, A Level Physics HQ, embedded from YouTube.

The two diagrams (the spectrum's shape is schematic; the line positions and energy levels are rounded literature values for molybdenum), the worked examples and the practice exercises are written by Ganzaa.

1A smooth hump with two sharp spikes

2The continuous spectrum: braking radiation

The Duane–Hunt limit

3The characteristic lines

Moseley's law

4Changing the tube settings

5How the spectrum is measured

A smooth hump with two sharp spikes

The continuous spectrum: braking radiation

The characteristic lines

Changing the tube settings

How the spectrum is measured