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Before anyone could measure an X-ray wavelength with a crystal, Charles Barkla was already sorting X-rays by their character, using little more than sheets of metal and an ionisation chamber. What he found, that each element sends out its own X-rays, became one of the main ways of studying atoms. This post tells his story from the Nobel archive and his own lecture, and works through the absorption and energy sums behind his method.
Who was Charles Barkla?
The Nobel Prize facts page records that he was born in Widnes, England, on 7 June 1877 and died in Edinburgh on 23 October 1944. His Nobel biography says he studied physics under Oliver Lodge at University College, Liverpool, worked with J. J. Thomson at the Cavendish Laboratory in Cambridge from 1899, and began his X-ray research in 1902. He held chairs in London from 1909 and in Edinburgh from 1913 until his death.
The 1917 prize summary explains an odd date. The committee decided that none of the 1917 nominations met the criteria, reserved the prize, and gave it to Barkla one year later, in 1918. The motivation was "for his discovery of the characteristic Röntgen radiation of the elements".
What did he find about scattered X-rays?
When X-rays pass through matter, the matter sends out secondary X-rays. Barkla showed that X-rays scattered at right angles to the beam are polarised, which he published in papers such as Polarisation in secondary Röntgen radiation (1906). In his Nobel lecture he compares it to "light scattered from the sky". The biography calls this a result of considerable importance, because it meant X-rays could be treated like ordinary light.
He also used scattering to count electrons. His lecture explains that the fraction of a beam lost by scattering depends on the number of electrons present, and that the most accurate data pointed to one electron per hydrogen atom, 6 for carbon, 7 for nitrogen and 8 for oxygen. Those are the atomic numbers, later confirmed by Rutherford, Bohr and Moseley.
What are characteristic X-rays?
The biography records that Barkla was the first to show secondary X-rays are of two kinds: X-rays scattered unchanged, and a fluorescent radiation peculiar to the substance. With Charles Sadler he reported homogeneous secondary Röntgen radiations in 1908. The Nobel facts page says that around 1906 he showed each element's secondary spectrum was unique, whatever its temperature, structure or chemical composition.
In his lecture Barkla lists the rules. Each element emits radiations characteristic of it. They are unaffected by its physical state or chemical combination. And only primary radiation of shorter wavelength can excite them. The radiations he found fell into two series, K and L; Siegbahn and colleagues later found an M series.
Worked example 1: is a beam homogeneous?
The sample readings below are invented for teaching. A beam passes through sheets of aluminium, each 0.10 mm thick. An ionisation chamber reads the intensity after each sheet.
0 sheets: 100. 1 sheet: 50. 2 sheets: 25. 3 sheets: 12.5.
Each sheet removes half. A constant fraction per layer means I = I₀e^(−μx). Half-thickness 0.10 mm = 0.010 cm gives μ = ln 2 / 0.010 ≈ 69 per cm. One value of μ describes the whole beam, so it behaves as one kind of radiation. That is the sense in which Barkla called his K radiation homogeneous.
Now a mixed beam: readings 100, 60, 40, 30. The fractions passed are 0.60, 0.67 and 0.75. The beam gets more penetrating as it goes, because the softer part is absorbed first. That is a sign of a mixture.
Worked example 2: what can excite iron's K radiation?
Barkla's rule says only shorter-wavelength primary X-rays can excite a characteristic line. The X-Ray Data Booklet from Lawrence Berkeley National Laboratory gives iron's K-shell binding energy as 7112 eV, and its emission table gives the Kα₁ line at 6403.84 eV. Use E = hc/λ with hc ≈ 1239.84 eV nm.
Primary beam at 0.20 nm. E = 1239.84/0.20 ≈ 6199 eV. Less than 7112 eV, so it cannot remove a K electron. No K radiation.
Primary beam at 0.15 nm. E ≈ 8266 eV. More than 7112 eV, so K radiation appears.
Threshold wavelength. λ = 1239.84/7112 ≈ 0.174 nm. Only primary X-rays shorter than this work.
The emitted line. λ = 1239.84/6403.84 ≈ 0.194 nm, longer than the 0.15 nm beam that excited it, as Barkla's rule requires.
Common mistakes with X-ray absorption questions
Subtracting a fixed amount per layer. Absorption removes a fixed fraction, so the decay is exponential, not linear.
Using the line energy as the threshold. Excitation needs the binding energy, which is higher than the emitted line's energy.
Mixing eV and keV. Write 7112 eV, not 7.112 eV.
Thinking the target's chemistry changes K lines. Barkla showed it does not.
The JEE (Advanced) 2026 syllabus lists characteristic and continuous X-rays, Moseley's law and the law of radioactive decay, which has the same exponential form. Check the current year's brochure. Our guide to X-rays and Moseley's law covers line energies. If the exponential step is the weak point, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing.
What can a student take from how Barkla worked?
Use the tool you have. He sorted X-rays by absorption years before crystal spectrometers.
Look for what stays the same. A line that ignores chemistry must belong to the atom.
Test a theory where it could fail. His scattering data checked electron numbers to see whether the wave picture held.
Practice routine: find μ if 0.20 mm halves a beam, find the fraction left after 1 mm, and find the threshold wavelength for copper's K shell from the data booklet.
Next step: take any absorption question, write I/I₀ after each layer, and check whether the fraction stays constant.
