# X-Rays and Compton Effect

> X-rays and Compton effect for JEE Physics: cutoff wavelength 12400/V, characteristic lines, Moseley's law and Compton shift 2.43 pm.

- Canonical URL: https://prepelephant.com/topics/jee/physics/x-rays-and-compton-effect
- Exam / course: JEE · Subject: Physics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "X-Rays and Compton Effect", PrepElephant, https://prepelephant.com/topics/jee/physics/x-rays-and-compton-effect

## Direct answer

When fast electrons decelerate in a metal target, they emit X-rays with a continuous spectrum (bremsstrahlung) cutting off sharply at lambda(min) = hc/(eV) = 12400/V angstrom with V in volts — 12.4 kV yields 1 angstrom — plus sharp characteristic lines (K alpha, K beta) whose frequencies obey Moseley's law, sqrt(nu) proportional to (Z − b). About 99% of the electron energy heats the target, only 1% becomes radiation. The Compton effect is the particle face: X-rays scattering off a loosely bound electron shift their wavelength by delta-lambda = (h/(m(e) c))(1 − cos theta), where h/(m(e) c) = 2.43 pm is the Compton wavelength, maximal (4.86 pm) at backscatter — evidence light delivers momentum in quanta.

## What you must remember

- **Cutoff wavelength:** lambda(min) = 12400/V(V) angstrom = 12.4 kV per angstrom in practical units; raising tube voltage shifts the entire continuum to shorter wavelengths and raises intensity.
- **Characteristic lines:** K alpha (L to K transition), K beta (M to K); Moseley's law sqrt(nu) = a(Z − b), with b = 1 for the K series — the linear anchor of the periodic table's nuclear charge.
- **Efficiency:** roughly 99% heat, 1% X-rays at a tungsten target; tungsten's high melting point and large Z earn the job, and rotating anodes spread the heat.
- **Compton shift formula:** delta-lambda = 2.43 × 10^-12 m × (1 − cos theta); zero at theta = 0, 2.43 pm at 90 degrees, 4.86 pm at 180 degrees.
- **What shifts and what doesn't:** wavelength depends only on scattering angle — not on the incident wavelength or the target material; photon energy drops by the recoil kinetic energy given to the electron.
- **Mechanism condition:** Compton scattering needs photons energetic enough (X-ray and above) to treat the electron as free; visible light on electrons gives the photoelectric regime instead.
- **Continuity with the photoelectric effect:** both demand photons, but photoelectric absorption transfers all energy to a bound electron while Compton scattering shares it with a recoil electron.
- **Pattern note:** Main computes lambda(min) and Compton shifts; Advanced asks Moseley-based identification of elements and energy–momentum bookkeeping of the recoil electron.

## Reading an X-ray spectrum like a story

The spectrum from a tube reads as two chapters. The smooth continuum from bremsstrahlung — electrons braking in the nucleus's field, each emitting a photon of whatever energy up to its full eV — ends at 12400/V: at 40 kV, lambda(min) = 0.31 angstrom is a wall whose position measures the accelerating voltage. Superimposed spikes appear only above each series' threshold voltage: K lines of tungsten need the K-shell electron knocked out first, and their wavelengths are the element's fingerprint. Moseley's law turns those fingerprints into element identification: if sqrt(nu) = a(Z − 1), measuring K alpha frequencies of successive elements exposes Z.

The Compton half is momentum arithmetic. At 90-degree scattering, delta-lambda = 2.43 pm regardless of whether the incident photon was 20 keV or 60 keV; the scattered photon always emerges 2.43 pm longer, and an incident wavelength much larger than the shift makes the effect invisible — which is why optical photons show no measurable Compton scattering.

## Where students slip

The cutoff wavelength belongs to the continuum, not the characteristic lines; raising V moves lambda(min) down but leaves K alpha's wavelength fixed (only its intensity rises once above threshold). Second, the Compton shift is independent of the incident wavelength — a point the exam tests by offering the tempting "shorter incident wavelength, bigger shift" distractor. Third, the 12400 carries units: with V in volts, lambda emerges in angstroms; feeding kilovolts without adjusting gives answers 1000 times wrong. Fourth, Moseley's b differs between series (1 for K, about 7.4 for L), so K-alpha and L-alpha lines are not parallel lines on the sqrt(nu) versus Z plot. And the Compton wavelength 2.43 pm is for the electron — other particles carry their own h/(mc).

## Frequently asked questions

### What determines the minimum wavelength in an X-ray spectrum?

The full conversion of one electron's kinetic energy eV into a single photon: lambda(min) = hc/(eV) = 12400/V angstrom, purely a function of tube voltage.

### What is Moseley's law and what did it fix?

K-series frequencies obey sqrt(nu) proportional to (Z − 1), giving a straight-line measure of atomic number that put the periodic table on nuclear-charge footing rather than atomic mass.

### What is the Compton shift at 90 degrees?

delta-lambda = h/(m(e) c) = 2.43 × 10^-12 m, since (1 − cos 90) = 1; the shift doubles to 4.86 pm at 180 degrees.

### Why is Compton scattering unobservable with visible light?

The shift 2.43 pm is a fixed fraction only near X-ray wavelengths — visible light's 500 nm wavelength changes by a relative 10^-5, unmeasurable, and bound-electron behaviour dominates.

### How are the photoelectric effect and Compton effect different?

Photoelectric absorption transfers the photon's entire energy to a bound electron; Compton scattering treats the electron as free and shares energy and momentum between scattered photon and recoil electron.