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Moseley's Law Characteristic K-Alpha X-Ray Energy & Atomic Number Z

Calculate characteristic K-alpha X-ray photon energy E_Ka = (3/4) * R_inf * (Z - 1)^2 (keV), wavelength lambda (Angstroms), and inverse elemental identification Z = 1 + SQRT(E_eV / 10.2043) for XRF/EDS spectroscopy.

Identify chemical elements from EDS/XRF K-alpha X-ray fluorescence peaks!

In 1913, Henry Moseley proved that periodic-table order reflects nuclear proton count Z rather than atomic weight: when a high-energy electron knocks out an inner K-shell (n = 1) electron and an L-shell (n = 2) electron drops down, the remaining single K-shell electron screens 1 proton charge (sigma_K = 1), giving Characteristic K-alpha Energy: E_Ka = (3 / 4) * 13.6057 eV * (Z - 1)^2 = 10.2043 * (Z - 1)^2 eV!

Dividing by 1,000 yields photon energy in keV, and dividing Planck-speed hc = 12.3984 keV·A by E_keV gives crystallographic X-ray tube wavelength in Angstroms (e.g., Copper Z = 29 gives Moseley E_Ka = 8.000 keV and lambda = 1.550 A, within 0.5% of relativistic fine-structure Cu K-alpha 8.048 keV / 1.541 A)! Inverting the law identifies an unknown EDS peak energy directly: Z = ROUND(1 + SQRT((E_keV * 1000) / 10.2043), 0)!

X-Ray Anode / XRF ElementAtomic Number ZMoseley K-alpha E (keV)Wavelength lambda = 12.3984 / E (A)
Copper XRD Tube (Cu)Z = 298.000 keV1.550 Angstrom
Molybdenum XRD Tube (Mo)Z = 4217.153 keV0.723 Angstrom
Tungsten CT Anode (W)Z = 7454.379 keV0.228 Angstrom

Moseley K-Alpha X-Ray Photon Energy (keV) from Atomic Number Z

Evaluates E_keV = (10.2043 * (Z - 1)^2) / 1000 for elements Z >= 2 (Helium and heavier).

=IF(B2 < 2, "Requires Atomic Number Z >= 2", ROUND((10.2043 * ((B2 - 1) ^ 2)) / 1000, 3))

Inverse EDS/XRF Elemental Atomic Number Z from Peak Energy (keV)

Inverts Moseley's law to recover integer atomic number Z from measured K-alpha peak energy B3 (keV).

=IF(B3 <= 0, "Invalid Photon Energy", ROUND(1 + SQRT((B3 * 1000) / 10.2043), 0))

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