X-Rays: Production, Spectra & Moseley's Law

X-rays became the workhorse of both medicine and materials science. Medical radiography and CT scanning use exactly this tube — bremsstrahlung from a spinning tungsten anode — to see through soft tissue to bone and to reconstruct three-dimensional slices of the body. The characteristic lines drive X-ray crystallography: shine monochromatic Cu $K_\alpha$ ($0.155\ \mathrm{nm}$) or Mo $K_\alpha$ ($0.0723\ \mathrm{nm}$) on a crystal and its diffraction pattern reveals the atomic arrangement — the method that solved the structure of DNA, haemoglobin, and countless drugs. Moseley's law lives on as X-ray fluorescence (XRF): bombard an unknown sample, read the characteristic line energies, and you have a non-destructive elemental analysis used everywhere from airport security and art authentication to sorting scrap metal and prospecting for ore. At the high end, synchrotron light sources push these ideas to extraordinary brightness for structural biology and semiconductor inspection. Every one of these rests on the same two facts this article establishes: a voltage-set continuous cutoff, and element-set characteristic lines that count the protons in the target.

An X-ray tube is a purpose-built collision. A filament is heated white-hot so that it boils off electrons (thermionic emission, from the electron-emission article). A large potential difference $V$ — anything from about $20\ \mathrm{kV}$ in a dental unit to $150\ \mathrm{kV}$ in a chest scanner — is applied between the filament (cathode) and a metal target (anode), so each electron is hurled across the vacuum and arrives at the target with a kinetic energy of exactly $eV$. When that electron smashes into the metal it stops, violently, and a decelerating charge radiates. Most of the energy just heats the target — which is why the anode is a heavy block of tungsten, often spinning and water-cooled, so it does not melt. But a small fraction comes off as X-rays: electromagnetic waves with wavelengths around $0.01$ to $10\ \mathrm{nm}$, thousands of times shorter than visible light.

Now look at what comes out. Spread the emerging X-rays by wavelength — using a crystal as a diffraction grating, since ordinary gratings are far too coarse — and the spectrum has two ingredients laid on top of each other. There is a broad, smooth hill of intensity spanning a wide range of wavelengths, and rising out of that hill are one or two needle-sharp peaks. The smooth hill is the continuous (or bremsstrahlung, German for "braking radiation") spectrum; the sharp peaks are the characteristic X-rays. They are made by two entirely different mechanisms, and — this is the crux of the whole topic — they respond to different knobs.

The continuous spectrum comes from bremsstrahlung. An incoming electron, plunging past the positive nuclei of the target atoms, is deflected and decelerated by their electric fields, and any accelerating (here, decelerating) charge radiates electromagnetic energy. Each electron loses its energy in one, two, or many small braking events, and can give up anywhere from a sliver to — in a single head-on stop — its entire kinetic energy. Because the energy lost per collision can take any value, the emitted photons take any energy up to a maximum, and the spectrum is smooth. The detailed shape of that hump (why it rises, peaks, and falls) needs the classical theory of radiation from accelerating charges and is beyond this article; we take the shape as observed and concentrate on the one feature that is exactly predictable with school algebra: its sharp edge on the short-wavelength side. That edge exists because there is a hardest possible photon. An electron cannot give a photon more energy than it has. The most energetic X-ray is born when a single electron dumps its entire kinetic energy $eV$ into one photon in one stop. That photon has the highest frequency $f_{\max}$ and therefore the shortest wavelength $\lambda_{\min}$:

where: $e = 1.60\times10^{-19}\ \mathrm{C}$ is the electron charge, $V$ the accelerating voltage in volts, $h = 6.63\times10^{-34}\ \mathrm{J\,s}$ Planck's constant, and $c = 3.00\times10^{8}\ \mathrm{m\,s^{-1}}$. The physical reading is a clean statement of energy conservation: the electron's whole purse of kinetic energy, all $eV$ of it, is spent buying one photon. Rearranging gives the Duane–Hunt limit:

where: the shortcut $hc/e = 1240\ \mathrm{eV\,nm}$ is the same one from the atomic-spectra work, and $V$ must be put in volts (not kilovolts) to land in nanometres. The astonishing content of eq (2) is what is missing from it: the target metal. Tungsten, molybdenum, copper — swap the anode and the short-wavelength cutoff does not budge. It depends on the accelerating voltage and on fundamental constants alone, which is exactly why measuring $\lambda_{\min}$ against $V$ is one of the cleanest tabletop ways to pin down Planck's constant $h$. Numerically, a tube run at $50\ \mathrm{kV}$ has $\lambda_{\min} = 1240/50000 = 0.0248\ \mathrm{nm}$, about a quarter of an ångström — a fifth of the size of an atom. Halve the voltage to $25\ \mathrm{kV}$ and the cutoff doubles to $0.0496\ \mathrm{nm}$: shorter wavelengths, harder X-rays, need higher voltage.

The sharp peaks are a different story entirely. When an incoming electron carries enough energy, it can knock a tightly bound electron clean out of an inner shell of a target atom — say the innermost $K$ shell ($n=1$). That leaves a vacancy, a hole in the atom's core, and the atom is now in a high-energy, unstable state. Almost instantly an electron from a higher shell drops down to fill the hole, and the energy it loses in the fall is emitted as a single photon. Because the inner energy levels of a given element are fixed and sharply defined, these photons come out at precise, discrete energies — the characteristic X-rays. A drop from the $L$ shell ($n=2$) into a $K$-shell hole gives the $K_\alpha$ line; a drop from the $M$ shell ($n=3$) into the same hole gives the (higher-energy) $K_\beta$ line. The wavelengths of these lines are a fingerprint: copper always emits its $K_\alpha$ at the same wavelength, tungsten at its own, and no two elements share the set. Two facts about the characteristic lines are exam gold. First, their positions depend on the target element only — change the metal and the lines jump; change the voltage and they stay put. Second, they appear only once the tube voltage exceeds a threshold: the incoming electron must carry at least enough energy to eject the inner-shell electron in the first place. Below that threshold you get the continuous spectrum but no characteristic lines. Above it, the lines appear at their fixed wavelengths and simply grow brighter. Why does the fingerprint change so regularly from element to element? Because the energy of an inner-shell transition is governed, to good approximation, by the same physics as the hydrogen atom — but with a nucleus of charge $Ze$ instead of $+e$. An electron dropping into a $K$-shell hole feels almost the full nuclear pull, screened by only the one electron remaining in that innermost shell, so it behaves as though the effective charge were $(Z-1)e$. That single insight, turned into a measurement by Henry Moseley in 1913, is the payoff of the whole topic.

Moseley photographed the $K_\alpha$ lines of dozens of elements and plotted the square root of each line's frequency against the element's position in the periodic table. The points fell on a dead-straight line. Written out, his result is

where: $\nu$ is the frequency of the characteristic line, $Z$ is the atomic number (the number of protons), $\sigma$ is a small screening constant ($\sigma \approx 1$ for the $K_\alpha$ line), and $a$ is a constant that is the same for every element on a given series. The physical reading: the square root of the line frequency counts up in equal steps as you step through the elements — one straight staircase, one step per proton. This is a statement about $Z$, the nuclear charge, and not about atomic weight. To see where eq (3) comes from and to get the working numbers, model the $K_\alpha$ transition as a one-electron fall from $n=2$ to $n=1$ around an effective charge $(Z-1)e$, using the Bohr energy formula from the sibling article, $E_n = -13.6\,Z_{\text{eff}}^2/n^2\ \mathrm{eV}$:

where: the factor $13.6\times(1 - \tfrac14) = 13.6\times\tfrac34 = 10.2\ \mathrm{eV}$ is the hydrogen $K_\alpha$-type energy scaled up by the square of the effective charge. Since $E = h\nu$, taking the square root of eq (4) gives $\sqrt{\nu} = \sqrt{E/h} \propto (Z-1)$ exactly as Moseley found — the $(Z-1)^2$ inside eq (4) is what makes $\sqrt{\nu}$, not $\nu$, the quantity that runs in a straight line.

The consequence was revolutionary. Before Moseley, the periodic table was ordered by atomic weight, and a handful of pairs sat awkwardly out of chemical order (argon is heavier than potassium, yet chemistry demands argon come first). Moseley's law gave each element an unambiguous integer $Z$ read straight off its X-ray line, resolving every such conflict, exposing gaps where undiscovered elements (like technetium and promethium) had to sit, and proving that the true ordering principle is nuclear charge — the number of protons — not mass. It also gave $Z$ a physical meaning it had never had: not just a position in a list, but a measurable count of the charge in the nucleus. Numerically, for copper ($Z=29$): $E_{K_\alpha} = 10.2\times(29-1)^2 = 10.2\times784 = 7997\ \mathrm{eV} \approx 8.0\ \mathrm{keV}$, a wavelength $\lambda = 1240/7997 = 0.155\ \mathrm{nm}$ — the standard "Cu $K_\alpha$" source used in X-ray crystallography labs the world over. For iron ($Z=26$): $10.2\times25^2 = 6375\ \mathrm{eV} \approx 6.4\ \mathrm{keV}$; for molybdenum ($Z=42$): $10.2\times41^2 = 17146\ \mathrm{eV} \approx 17.1\ \mathrm{keV}$.

Board level — the short-wavelength limit (ICSE and JEE Main pattern)

An X-ray tube operates at $50\ \mathrm{kV}$. Taking $h = 6.63\times10^{-34}\ \mathrm{J\,s}$, $c = 3.0\times10^{8}\ \mathrm{m\,s^{-1}}$, $e = 1.60\times10^{-19}\ \mathrm{C}$ and $hc = 1240\ \mathrm{eV\,nm}$, find (a) the maximum energy of an emitted X-ray photon, (b) the short-wavelength limit $\lambda_{\min}$, and (c) the new $\lambda_{\min}$ if the voltage is lowered to $25\ \mathrm{kV}$.

Step 1. Each electron reaches the target with kinetic energy $eV = 50\ \mathrm{keV}$, and the hardest photon carries all of it: $E_{\max} = eV = 50\ \mathrm{keV} = (1.60\times10^{-19})(5.0\times10^{4}) = \boxed{8.0\times10^{-15}\ \mathrm{J}}$.

Step 2. Short-wavelength limit from eq (2): $\lambda_{\min} = 1240/V = 1240/50000 = \boxed{0.0248\ \mathrm{nm}}$ (that is $24.8\ \mathrm{pm}$, about a quarter of an ångström). Remember to put $V$ in volts, not kilovolts, to get nanometres.

Step 3. At $25\ \mathrm{kV}$, $\lambda_{\min} = 1240/25000 = \boxed{0.0496\ \mathrm{nm}}$. Halving the voltage doubles the cutoff wavelength — the inverse relationship $\lambda_{\min} \propto 1/V$ in action.

Board level — reading a characteristic line (ICSE pattern)

The $K_\alpha$ line of a target is found at wavelength $0.155\ \mathrm{nm}$. Using $hc = 1240\ \mathrm{eV\,nm}$ and $E_{K_\alpha} = 10.2(Z-1)^2\ \mathrm{eV}$, find the photon energy of the line and the atomic number of the target metal.

Step 1. Photon energy: $E = hc/\lambda = 1240/0.155 = \boxed{8000\ \mathrm{eV} = 8.0\ \mathrm{keV}}$.

Step 2. Invert Moseley's energy form: $(Z-1)^2 = E/10.2 = 8000/10.2 = 784$, so $Z - 1 = \sqrt{784} = 28$ and $Z = \boxed{29}$ — copper.

Competitive level — Moseley identifies an unknown (JEE Advanced pattern)

The $K_\alpha$ characteristic X-ray from an unknown metal target has energy $6.4\ \mathrm{keV}$. Using the screened-Bohr result $E_{K_\alpha} = 10.2(Z-1)^2\ \mathrm{eV}$, identify the element, and find the $K_\alpha$ energy you would expect from molybdenum ($Z = 42$).

Step 1. Solve for $Z$: $(Z-1)^2 = E/10.2 = 6400/10.2 = 627$, so $Z - 1 = \sqrt{627} = 25.0$ and $Z = \boxed{26}$ — iron. (Iron's real $K_\alpha$ sits at $6.40\ \mathrm{keV}$, so the model nails it.)

Step 2. Molybdenum, $Z = 42$: $E_{K_\alpha} = 10.2\times(42-1)^2 = 10.2\times41^2 = 10.2\times1681 = 17146\ \mathrm{eV} \approx \boxed{17.1\ \mathrm{keV}}$, a wavelength $\lambda = 1240/17146 = 0.0723\ \mathrm{nm}$. Molybdenum's harder $K_\alpha$ is why it is the standard source for small-molecule crystallography, where you want to reach shorter wavelengths than copper offers. Timing note: about 75 seconds. The whole item is two uses of $E = 10.2(Z-1)^2$ — once inverted to get $Z$, once forward for molybdenum. The trap is arithmetic: keep energies in eV inside the formula ($6400$, not $6.4$) and take the square root cleanly. Recognising that $\sqrt{627} \approx 25$ gives iron by inspection is worth the mark.

Competitive level — Planck's constant from the cutoff (JEE Advanced pattern)

When an X-ray tube is run at $50\ \mathrm{kV}$ the short-wavelength limit is measured as $0.0248\ \mathrm{nm}$. Treating this as a measurement, recover Planck's constant $h$ from $eV = hc/\lambda_{\min}$, using $e = 1.60\times10^{-19}\ \mathrm{C}$ and $c = 3.0\times10^{8}\ \mathrm{m\,s^{-1}}$.

Step 1. Rearrange eq (1) for $h$: $h = eV\lambda_{\min}/c$.

Step 2. Substitute: $h = (1.60\times10^{-19})(5.0\times10^{4})(0.0248\times10^{-9})/(3.0\times10^{8})$. The numerator is $(8.0\times10^{-15})(2.48\times10^{-11}) = 1.98\times10^{-25}$, and dividing by $3.0\times10^{8}$ gives $h = \boxed{6.6\times10^{-34}\ \mathrm{J\,s}}$ — Planck's constant, recovered from an X-ray cutoff to two figures. Timing note: about 90 seconds. This is the Duane–Hunt method of measuring $h$, historically important and a favourite of examiners precisely because it inverts the usual direction — you are given $\lambda_{\min}$ and asked for $h$. The arithmetic is one product over one division; carry the powers of ten carefully.

The cutoff-versus-lines swap. The number-one error: thinking the characteristic lines shift with voltage or the continuous cutoff depends on the target. It is the other way round — $\lambda_{\min} = 1240/V$ (voltage only); $K_\alpha, K_\beta$ (target only). Any question that changes one variable is testing whether you keep these separate. Appears on ICSE, JEE Main and Advanced, and A-Level 9702.

Volts, not kilovolts, in $1240/V$. The shortcut $\lambda_{\min}[\mathrm{nm}] = 1240/V$ needs $V$ in volts. A $50\ \mathrm{kV}$ tube is $V = 50000$, giving $0.0248\ \mathrm{nm}$; write $1240/50$ and you get a nonsensical $24.8\ \mathrm{nm}$ (ultraviolet, not X-ray). Always sanity-check that an X-ray wavelength lands around $0.01$–$0.1\ \mathrm{nm}$.

Moseley uses $(Z-1)$, not $Z$, for $K_\alpha$. The screening constant $\sigma \approx 1$ for the $K_\alpha$ line is not optional — one inner electron shields the nucleus. Using $Z$ instead of $(Z-1)$ throws copper's line off by about $7\%$ and loses the mark on any "find the element" problem. (For the $L$ series the screening is larger, $\sigma \approx 7.4$, but school problems stay with $K_\alpha$.)

Keep energies in eV inside $10.2(Z-1)^2$. The coefficient $10.2$ is in electronvolts, so a $6.4\ \mathrm{keV}$ line goes in as $6400$, not $6.4$. Mixing units here is the most common arithmetic slip when solving Moseley problems for $Z$.

$\sqrt{\nu}$ is linear in $Z$, not $\nu$. Moseley's straight line is $\sqrt{\nu}$ against $Z$, because the energy goes as $(Z-1)^2$. A question asking "what is linear in atomic number?" wants $\sqrt{\nu}$ (or $\sqrt{E}$), not the frequency or the wavelength itself.

Board scope — do not claim this for CBSE or NEET. X-ray production spectra and Moseley's law are ICSE, JEE Main, JEE Advanced and A-Level (CIE 9702) material. They are not in the core CBSE Class 12 syllabus and not on NEET, and X-ray/Moseley content is not in AP Physics 2 either. If you sit CBSE or NEET, this page is enrichment, not a required topic.

Texts. ISC Physics Class 12, the chapter on atoms and X-rays · Irodov / D. C. Pandey Modern Physics for the Moseley and Duane–Hunt problem sets · Halliday, Resnick and Walker, "All About Atoms", the section on X-rays and Moseley's law · Eisberg and Resnick, Quantum Physics, for the characteristic-spectrum treatment beyond the HS ceiling.

Provenance. Representative constants and standard exam values. Recompute: $\lambda_{\min}(50\,\mathrm{kV}) = 1240/50000 = 0.0248\ \mathrm{nm}$; $\lambda_{\min}(25\,\mathrm{kV}) = 0.0496\ \mathrm{nm}$; $eV(50\,\mathrm{kV}) = 8.0\times10^{-15}\ \mathrm{J}$; $f_{\max} = 1.21\times10^{19}\ \mathrm{Hz}$; Moseley coefficient $13.6\times\tfrac34 = 10.2\ \mathrm{eV}$; Cu $K_\alpha = 10.2\times28^2 = 7997\ \mathrm{eV} \approx 8.0\ \mathrm{keV}$, $\lambda = 0.155\ \mathrm{nm}$; Fe $K_\alpha = 10.2\times25^2 = 6375\ \mathrm{eV} \approx 6.4\ \mathrm{keV}$ ($Z = 26$); Mo $K_\alpha = 10.2\times41^2 = 17146\ \mathrm{eV} \approx 17.1\ \mathrm{keV}$, $\lambda = 0.0723\ \mathrm{nm}$; Moseley slope $a = \sqrt{10.2/(4.14\times10^{-15})} = 4.96\times10^{7}\ \sqrt{\mathrm{Hz}}$; $h$ recovered from cutoff $= 6.6\times10^{-34}\ \mathrm{J\,s}$.

Prerequisites. Atomic Spectra: Emission & Absorption Lines — where line spectra as atomic fingerprints and the photon relation $E = hf$ are established; the characteristic X-ray lines are the inner-shell version of the same idea. The screened-Bohr estimate here also leans on the Bohr energy formula $E_n = -13.6\,Z^2/n^2\ \mathrm{eV}$ from The Bohr Model of the Atom (a sibling article, referenced in prose). Leads to. Hydrogen Spectrum Series (university seam) — where the series structure and Rydberg formula behind these transition energies are developed in full; treat it as a go-deeper pointer, never an exam dependency here.