X-ray Production and Energy Spectrum Revision Notes | A Level AQA Physics

1. How X-rays Are Born: The Big Picture

Picture a beam of electrons, each carrying a precise amount of kinetic energy, fired at a dense metal target at enormous speed. When those electrons slam into the target, their energy has to go somewhere — and a significant fraction of it leaves as X-ray photons. That conversion process, and the spectrum of photon energies it produces, is exactly what this lesson unpacks.

X-ray production sits at the intersection of two ideas you already know well: the work done on a charge by a potential difference, and the photon model of electromagnetic radiation. Bringing those two ideas together gives you a complete picture of the X-ray energy spectrum.

2. Inside the X-ray Tube: Physical Principles

An X-ray tube contains a heated cathode filament and a metal anode target, held under high vacuum. The filament is heated until thermionic emission occurs — electrons boil off the surface. A large potential difference $V$ (the accelerating voltage, typically 20–150 kV in medical imaging) is applied between cathode and anode, accelerating those electrons across the gap.

Each electron gains kinetic energy equal to the work done on it by the electric field. Because the electron carries charge $e$ and is accelerated through potential difference $V$, its kinetic energy on arrival at the anode is $E_k = eV$. This is the energy budget available to produce X-ray photons — and it sets a hard ceiling on the photon energies that can be emitted.

When the high-speed electrons reach the anode, two distinct physical processes produce X-ray photons. The first is a broad, continuous process; the second produces sharp, discrete peaks. Understanding both is essential for interpreting the X-ray energy spectrum.

2a. The Continuous Spectrum: Bremsstrahlung

Most incoming electrons do not stop in a single collision. Instead, each electron passes close to one or more atomic nuclei in the target material and is deflected — decelerated — by the electrostatic (Coulomb) interaction with atomic nuclei. Each deceleration event converts some of the electron's kinetic energy into a single X-ray photon. The German term for this is Bremsstrahlung, meaning 'braking radiation'.

Because each electron undergoes a different degree of deceleration in a different interaction, the photons produced carry a continuous range of energies from near-zero up to a definite maximum. The result is a smooth, continuous spectrum — sometimes called the white radiation or continuous background.

The maximum photon energy in this continuous spectrum corresponds to the rare event where an electron loses all of its kinetic energy in a single interaction, producing one photon of energy $E_{max} = eV$. This sets the short-wavelength cut-off of the spectrum: no photon can have more energy than the electron that produced it.

Because photon energy and wavelength are related by $E = \frac{hc}{\lambda}$, the maximum photon energy corresponds to a minimum wavelength $\lambda_{min}$. Rearranging gives the short-wavelength limit of the spectrum.

2b. The Characteristic Spectrum: Discrete Peaks

Superimposed on the continuous bremsstrahlung background are sharp, narrow peaks at specific photon energies. These are the characteristic X-rays, and they arise from a completely different physical process — one that involves the internal energy-level structure of the target atoms.

When an incoming electron has enough kinetic energy, it can knock an inner-shell electron (for example, a K-shell electron) completely out of a target atom, leaving a vacancy. An electron from a higher energy shell — say the L or M shell — then drops down to fill that vacancy. The energy released in this transition is emitted as a single X-ray photon whose energy equals the difference between the two shell energies: $E_{photon} = E_{upper} - E_{lower}$.

Because atomic energy levels are fixed and unique to each element, the photon energies produced are fixed and unique to the target material. This is why the peaks are called characteristic — they are a fingerprint of the anode element. Change the target material and the peak positions shift; change the voltage and the peak positions stay the same (provided the voltage is still high enough to cause ionisation of the inner shells).

3. The Complete X-ray Energy Spectrum

Putting both processes together gives the full X-ray spectrum: a broad, continuous bremsstrahlung curve with sharp characteristic peaks rising above it at specific energies. The chart below shows the key features you need to be able to describe and explain in an exam.

Notice the sharp cut-off at the high-energy (short-wavelength) end of the continuous spectrum. That cut-off sits at exactly $E_{max} = eV$ — it does not tail off gradually. The curve rises from zero at $E_{max}$, reaches a broad peak at roughly one-third to one-half of $E_{max}$, then falls back towards zero at very low photon energies (very low-energy photons are absorbed within the tube itself before they can exit).

The characteristic peaks sit at fixed positions determined by the target material. They only appear if the accelerating voltage is high enough to ionise the relevant inner shell — below that threshold voltage, the peaks are absent entirely.

4. Calculating Maximum Photon Energy and Minimum Wavelength

Exam questions on this topic almost always ask you to calculate either the maximum photon energy or the minimum wavelength for a given accelerating voltage. The method is straightforward, but unit handling is where marks are lost. Work through the example below carefully — pay particular attention to the unit conversion for voltage.

Worked example: Calculating Maximum Photon Energy and Minimum Wavelength

An X-ray tube operates with an accelerating voltage of 80 kV. Calculate (a) the maximum photon energy and (b) the minimum wavelength of X-rays produced. ($h = 6.63 \times 10^{-34}$ J s, $c = 3.0 \times 10^{8}$ m/s, $e = 1.6 \times 10^{-19}$ C)

1. $V = 80\text{ kV} = 80 \times 10^{3}\text{ V} = 8.0 \times 10^{4}\text{ V}$ — Always convert kV to V before substituting. This is the most common source of errors in this calculation. [M1]
2. $E_{max} = 1.6 \times 10^{-19} \times 8.0 \times 10^{4}$ — Multiply the elementary charge by the accelerating voltage in volts. [M1]
3. $E_{max} = 1.6 \times 8.0 \times 10^{-19+4} = 12.8 \times 10^{-15} = 1.28 \times 10^{-14}\text{ J}$ — Multiply coefficients: $1.6 \times 8.0 = 12.8$. Add exponents: $-19 + 4 = -15$. Then write in standard form. [A1]
4. $\lambda_{min} = \frac{hc}{E_{max}}$ — Since $E = \frac{hc}{\lambda}$, rearranging gives $\lambda = \frac{hc}{E}$. Use $E_{max}$ to get $\lambda_{min}$. [M1]
5. $\lambda_{min} = \frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{1.28 \times 10^{-14}}$ — Substitute $h$, $c$, and the $E_{max}$ found in step 3. Keep all values in SI units. [M1]
6. $hc = 6.63 \times 3.0 \times 10^{-34+8} = 19.89 \times 10^{-26} = 1.989 \times 10^{-25}\text{ J m}$ — Multiply coefficients: $6.63 \times 3.0 = 19.89$. Add exponents: $-34 + 8 = -26$.
7. $\lambda_{min} = \frac{1.989 \times 10^{-25}}{1.28 \times 10^{-14}} = \frac{1.989}{1.28} \times 10^{-25-(-14)} = 1.554 \times 10^{-11}\text{ m}$ — Divide coefficients: $1.989 \div 1.28 \approx 1.55$. Subtract exponents: $-25 - (-14) = -11$. [A1]
8. $\lambda_{min} \approx 1.55 \times 10^{-11}\text{ m} = 0.0155\text{ nm}$ — X-ray wavelengths are typically $10^{-11}$ to $10^{-8}$ m (0.01–10 nm). This value sits at the hard end of the X-ray range, consistent with a high accelerating voltage. Units are metres — correct for wavelength.

Answer: (a) $E_{max} = 1.28 \times 10^{-14}$ J; (b) $\lambda_{min} \approx 1.55 \times 10^{-11}$ m

5. How Operating Conditions Alter the Spectrum

Now that you can calculate the key quantities, it is worth stepping back to see how changing the tube's operating conditions reshapes the spectrum as a whole. Examiners frequently ask you to sketch or describe the effect of changing voltage or target material — so build a clear mental model here.

Increasing the accelerating voltage does three things simultaneously: it shifts $\lambda_{min}$ to a shorter value (higher $E_{max}$), it increases the overall intensity of the continuous spectrum (more energetic electrons produce more photons per unit time), and it raises the intensity of the characteristic peaks (more electrons exceed the ionisation threshold). The shape of the continuous curve broadens and the peak shifts upward in energy.

Changing the target material leaves the continuous spectrum shape broadly similar but moves the characteristic peak positions to entirely new energies, because those peaks depend solely on the atomic energy-level structure of the new element. A tungsten anode and a molybdenum anode operated at the same voltage produce the same $\lambda_{min}$ and a similar bremsstrahlung curve, but their characteristic peaks appear at completely different photon energies.

6. Exam Technique and Common Mistakes

Questions on X-ray production tend to follow predictable patterns. The most reliable way to pick up every available mark is to be precise about which process you are describing and to handle units carefully in calculations. The guidance below consolidates the key exam strategies for this topic.

For a 'describe and explain' question worth 4–6 marks, structure your answer by addressing the continuous spectrum and the characteristic spectrum separately. For each, name the process, describe the physical mechanism, and link it to the shape of the spectrum. Examiners award marks for the mechanism, not for restating the observation.

Together, the continuous bremsstrahlung spectrum and the characteristic peaks form the complete X-ray energy spectrum — a direct consequence of the physical principles of electron deceleration and atomic energy-level transitions, both governed by the conservation of energy and the photon model of electromagnetic radiation.