X-ray SpectraJEE Advanced

Cutoff wavelength & Moseley's law — interactive Physics simulation for IIT-JEE.

Concept

An X-ray tube fires keV electrons at a metal target. Deceleration radiates a continuous spectrum (bremsstrahlung) with a sharp cutoff: no photon can exceed the electron's full energy, so $\lambda_{min} = hc/eV$ — target-independent. Knock out a K-shell electron and sharp characteristic lines appear, whose frequencies follow Moseley's law $\sqrt\nu \propto (Z-1)$ .

Key formula

\lambda_{min} = \frac{hc}{eV} = \frac{12.4}{V(\text{kV})}\,\text{Å}, \qquad \sqrt{\nu_{K\alpha}} = a(Z - 1)

Derivation

Cutoff: the most extreme collision converts ALL kinetic energy eV into one photon: $eV = hc/\lambda_{min}$ .

Moseley: the L→K transition sees the nucleus screened by the one remaining K electron, effective charge (Z−1): $\nu = 2.47\times10^{15}(Z-1)^2$ Hz — a hydrogen-like Bohr formula. This ordered the periodic table by Z, not mass.

Scenarios to explore

X-ray Spectra — Cutoff λ_min = hc/eV and Moseley's (Z−1)² law.

Real-world applications

Medical radiography & CT (tube voltage sets penetration).
X-ray crystallography (needs sharp Kα lines).
XRF element identification via characteristic lines.

JEE exam tips

λ_min(Å) = 12.4/V(kV) — memorize the 12.4 shortcut (or 1240 eV·nm).
Doubling V halves λ_min; characteristic lines DON'T move (fixed by Z).
Kα appears only if eV exceeds the K-shell binding energy ≈ 13.6(Z−1)² eV.

Common mistakes

Thinking λ_min depends on the target material — it depends ONLY on voltage.
Moseley with Z instead of (Z−1).
Confusing continuous (bremsstrahlung) and characteristic (transition) origins.

Exam traps to avoid

Raising the filament current raises INTENSITY, not photon energy.
Characteristic lines sit ON TOP of the continuous hump, always at λ > λ_min.