11 min readConcept sheetSection A MCQ (1 mark) and objective (2 marks, two parts); Section B short answer (3 marks) or long answer (5 marks) with a labelled diagram; also a JEE favourite (Moseley's law)

X-rays: production, types and properties

Fast electrons suddenly stopped by a heavy metal target give X-rays: a continuous spectrum starting at λmin plus sharp characteristic lines set by the target.

Duane-Hunt laweV = hν(max) = hc/λmin, where V is the voltage across the tube

Moseley's law (Kα lines)ν = Rc(1/1² − 1/2²)(Z − 1)², with Rc = 3.29 × 10¹⁵ Hz (your NIOS lesson writes this constant simply as R); so √ν ∝ (Z − 1), where Z is the atomic number of the target

What controls whatFilament current controls the intensity; tube voltage (10 kV to 1 MV) controls the quality (penetrating power)

NatureElectromagnetic waves of very short wavelength; no charge, so fields do not deflect them

How X-rays are produced

Electrons moving at high speed give out X-rays when a heavy metal target brings them to a sudden stop.

1
A filament F (the cathode) is heated by a current. It gives out electrons by thermionic emission (emission by heating).
2
A high voltage, 10 kV to 1 MV, between the filament and the target speeds the electrons up.
3
The tube is highly evacuated, so the electrons reach the target without hitting gas molecules.
4
The target is a metal of high atomic number and high melting point, such as tungsten.
5
When the electrons strike the target and stop suddenly, X-rays come out. Only a small part of their energy, usually less than 1 per cent, becomes X-rays. The rest becomes heat, carried away by cold circulating water. (Your NIOS lesson gives about 5 per cent, a figure reached only at very high voltages.)
6★
The filament current controls the intensity (how many X-rays). The tube voltage controls the quality (how energetic and penetrating they are).

★ key step

Properties of X-rays

They are electromagnetic waves of very short wavelength and travel at the speed of light.
They blacken a photographic plate.
They make some chemicals fluoresce (glow), such as zinc sulphide.
They ionise gases they pass through.
They are not deflected by electric or magnetic fields, because they carry no charge.
Mirrors and glass do not reflect or refract them in the usual way, but the closely spaced atomic layers of crystals diffract them.
They pass easily through flesh, paper and wood, but bone and lead stop them much more.

Continuous and characteristic X-rays

Continuous X-raysMade as electrons are slowed down in collisions with target atoms. They have all wavelengths above a sharp lower limit λmin, which depends only on the tube voltage. A higher voltage raises the intensity at every wavelength and shifts the peak to shorter wavelengths.

Characteristic X-raysSharp lines (such as Kα and Kβ) standing on the continuous spectrum. Their wavelengths depend only on the target metal, not on the voltage. Each element has its own lines, like a fingerprint.

Moseley studied the characteristic lines of many elements and found that their frequency rises steadily with atomic number Z.

Worked example: shortest wavelength from a 20 kV tube

Problem

An X-ray tube works at 20 kV. Find the highest frequency and the shortest wavelength of the X-rays it gives out. (h = 6.63 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C, c = 3 × 10⁸ m/s)

Working

Duane-Hunt law: eV = hν(max) = hc/λmin.
Energy of each electron: eV = 1.6 × 10⁻¹⁹ C × 2 × 10⁴ V = 3.2 × 10⁻¹⁵ J.
ν(max) = eV/h = 3.2 × 10⁻¹⁵ ÷ 6.63 × 10⁻³⁴ = 4.83 × 10¹⁸ Hz.
λmin = c/ν(max) = 3 × 10⁸ ÷ 4.83 × 10¹⁸ = 6.2 × 10⁻¹¹ m = 0.62 Å.

Answer

ν(max) ≈ 4.83 × 10¹⁸ Hz; λmin ≈ 6.2 × 10⁻¹¹ m (0.62 Å)

Worked example: Moseley's law for copper

Problem

Using Moseley's law, find the frequency and wavelength of the Kα line of copper (Z = 29). (R = 1.097 × 10⁷ m⁻¹, c = 3 × 10⁸ m/s)

Working

Moseley's law for Kα: ν = Rc(1/1² − 1/2²)(Z − 1)² = (3/4)Rc(Z − 1)².
Rc = 1.097 × 10⁷ m⁻¹ × 3 × 10⁸ m/s = 3.291 × 10¹⁵ Hz.
(Z − 1)² = 28² = 784.
ν = 0.75 × 3.291 × 10¹⁵ × 784 = 1.94 × 10¹⁸ Hz.
λ = c/ν = 3 × 10⁸ ÷ 1.94 × 10¹⁸ = 1.55 × 10⁻¹⁰ m = 1.55 Å.

Answer

ν ≈ 1.94 × 10¹⁸ Hz; λ ≈ 1.55 × 10⁻¹⁰ m (1.55 Å)

How to draw the X-ray tube and its spectrum

Draw a long glass tube and write highly evacuated tube on it.
On the right, draw a small coil labelled filament F (cathode), joined to a low-voltage supply that heats it.
On the left, draw a slanted metal block labelled target (tungsten), with cooling water pipes behind it.
Join the filament (−) and target (+) to a high-voltage source. Draw arrows for electrons from F to the target and wavy arrows labelled X-rays leaving the target.
For the spectrum: draw intensity against wavelength. Draw a smooth hump starting sharply at λmin, with two tall, narrow peaks labelled Kα and Kβ on it.

In real life

Doctors use X-rays to see broken bones and dentists to find tooth decay. Airport baggage scanners use them too, and factories use them to find hidden cracks in metal castings. Lead aprons protect patients and staff.

Exam angle

3 marks: “Explain, with a labelled sketch of the tube, how X-rays are produced.” A heated filament gives electrons; a high voltage accelerates them in a highly evacuated tube; they strike a heavy, high-melting-point target (tungsten) and stop suddenly, giving X-rays; cooling water removes the heat. Diagram: filament, target, high-voltage supply, cooling, X-ray beam.

2 marks: “Distinguish between continuous and characteristic X-rays.” Continuous: all wavelengths above λmin; λmin depends only on the tube voltage. Characteristic: sharp lines at wavelengths fixed by the target element; they do not depend on the voltage.

2 marks: “An X-ray tube is run at 50 kV. Find the shortest wavelength emitted. (h = 6.63 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C, c = 3 × 10⁸ m/s)” λmin = hc/eV = (6.63 × 10⁻³⁴ × 3 × 10⁸) ÷ (1.6 × 10⁻¹⁹ × 5 × 10⁴) ≈ 2.5 × 10⁻¹¹ m (0.25 Å).

Remember it

Current counts, voltage punches: filament current sets how many X-rays; tube voltage sets how penetrating, with λmin = hc/eV.