X-Ray Tube (OCR A Level Physics)

Revision Note

Test Yourself
Katie M


Katie M



Structure of an X-ray Tube

  • An X-ray tube is a device that converts an electrical input into X-rays
  • It is composed of four main components:
    • A heated cathode
    • An anode
    • A metal target
    • A high voltage power supply


  • The production of X-rays has many practical uses, such as in:
    • Medical imaging (radiography)
    • Security
    • Industrial imaging


The main components of an X-ray Tube are the heated cathode, anode, metal target and a high voltage supply

The Role of the Components

Heated Cathode

  • At one end of the tube is the cathode (negative terminal) which is heated by an electric current
    • The heat causes electrons to be liberated from the cathode, gathering in a cloud near its surface
    • This process of thermionic emission is the source of the electrons 


  • At the other end of the tube, an anode (positive terminal) is connected to the high voltage supply
  • This allows the electrons to be accelerated up to a voltage of 200 kV 
    • When the electron arrives at the anode, its kinetic energy is 200 keV (by the definition of an electronvolt)
  • Only about 1% of the kinetic energy is converted to X-rays
    • The rest is converted to heat energy
    • Therefore, to avoid overheating, the anode is spun at 3000 rpm and sometimes water-cooled

Metal Target

  • When the electrons hit the target at high speed, they lose some of their kinetic energy
    • This is emitted as X-ray photons
  • A heat-resistant block of metal, usually Tungsten, is embedded at the end of the anode, facing the cathode
    • This is the material that the electrons collide with and X-rays are generated in

High Voltage Power Supply

  • The high voltage supply creates a large potential difference (> 50 kV) between the cathode and the target
    • This causes electrons in the cloud around the cathode to be accelerated to a high velocity towards the target, which they strike, creating X-rays

Other Components

  • X-rays are produced in all directions, so the tube is surrounded by lead shielding
    • This is to ensure the safety of the operators and recipients of the X-rays
    • An adjustable window allows a concentrated beam of X-rays to escape and be controlled safely
  • The anode and cathode are housed inside a vacuum chamber
    • This is to ensure that the electrons do not collide with any particles on their way to the metal target

Production of X-ray Photons

  • When the fast-moving electrons collide with the target, X-rays are produced by one of two methods
    • Method 1: Bremsstrahlung
    • Method 2: Characteristic Radiation

Method 1: Bremsstrahlung

  • When the high-speed electrons collide with the metal target, they undergo a steep deceleration
    • When a charged particle decelerates quickly, some of the energy released is converted into a photon
  • A small amount of the kinetic energy (~ 1%) from the incoming electrons is converted into X-rays as the electrons decelerate in the tungsten, due to conservation of energy 
    • The rest of the energy heats up the anode, which usually requires some form of cooling
  • The energy of the X-ray photon can be of any value, up to the original kinetic energy of the electron, giving a spread of possible X-ray energies
    • These X-rays cause the continuous or ‘smooth hump shaped’ line on an intensity wavelength graph


  • When an electron is accelerated, it gains energy equal to the electronvolt, this energy can be calculated using:

Emax = eV

  • This is the maximum energy that an X-ray photon can have
  • The smallest possible wavelength is equivalent to the highest possible frequency and therefore, the highest possible energy
    • This is assuming all of the electron’s kinetic energy has turned into electromagnetic energy
  • Therefore, the maximum X-ray frequency fmax, or the minimum wavelength λmin, that can be produced is calculated using the equation:

E subscript m a x end subscript equals e V equals h f subscript m a x end subscript equals fraction numerator h c over denominator lambda subscript m i n end subscript end fraction

  • The maximum X-ray frequency, fmax, is therefore equal to:

f subscript m a x end subscript equals fraction numerator e V over denominator h end fraction

  • The minimum X-ray wavelength, λmin, is therefore equal to:

lambda subscript m i n end subscript equals fraction numerator h c over denominator e V end fraction

  • Where:
    • e = elementary charge (C)
    • V = potential difference between the anode and cathode (V)
    • h = Planck's constant (J s)
    • c = the speed of light (m s−1)

Method 2: Characteristic Radiation

  • Some of the incoming fast electrons cause inner shell electrons of the tungsten to be ‘knocked out’ of the atom, leaving a vacancy
    • This vacancy is filled by an outer electron moving down and releasing an X-ray photon as it does (equal in energy to the difference between the two energy levels)
    • Because these X-rays are caused by energy level transitions, they have only specific discrete energies
    • They cause sharp spikes on an intensity wavelength graph
    • The number of spikes depends on the element used for the target - there are two sets of spikes for a tungsten target, representing two sets of possible energy transitions



Worked example

X-rays are a type of electromagnetic wave with wavelengths in the range 10−8 to 10−13 m

If the accelerating potential difference in an X-ray tube is 60 kV, determine if the photons emitted fall within this range.

Step 1: Write out known quantities

    • Charge on an electron, e = 1.6 × 10−19 C
    • Accelerating potential difference, V = 60 000 V
    • Planck’s constant, h = 6.63 × 1034 J s
    • Speed of light, c = 3 × 108 m s1


Step 2: Determine the maximum possible energy of a photon

    • The maximum possible energy of a photon corresponds to the maximum energy an electron could have:

Emax = eV

Step 3: Determine an expression for minimum wavelength

Planck relation:    E = hf

Wave equation:    c =

    • When energy is a maximum:

Emax = eV = hfmax

    • Maximum energy corresponds to a minimum wavelength:

e V equals fraction numerator h c over denominator lambda subscript m i n end subscript end fraction

    • Rearrange for minimum wavelength, λmin:

lambda subscript m i n end subscript equals fraction numerator h c over denominator e V end fraction

Step 4: Calculate the minimum wavelength λmin

lambda subscript m i n end subscript equals fraction numerator open parentheses 6.63 cross times 10 to the power of negative 34 end exponent close parentheses open parentheses 3 cross times 10 to the power of 8 close parentheses over denominator open parentheses 1.6 cross times 10 to the power of negative 19 end exponent close parentheses open parentheses 60 space 000 close parentheses end fraction

λmin = 2.1 × 1011 m

Step 5: Comment on whether this is within the range for the wavelength of an X-ray 

    • X-ray wavelengths are within 10−8 to 10−13 m
    • The minimum wavelength for a 60 kV supply is 2.1 × 1011 m, which means the photons produced will be X-rays

Worked example

A typical spectrum of the X-ray radiation produced by electron bombardment of a metal target is shown below.

Explain why:

A continuous spectrum of wavelengths is produced.
The gradient is steeper at shorter wavelengths.
The spectrum has a sharp cut-off at short wavelengths.

Part (a)

Step 1: Consider the path of the electrons from the cathode to the anode

    • Photons are produced whenever a charged particle undergoes a large acceleration or deceleration
    • X-ray tubes fire high-speed electrons at a metal target
    • When an electron collides with the metal target, it loses energy in the form of an X-ray photon as it decelerates

Step 2: Consider the relationship between the energy of the electron and the wavelength of the photon

    • The wavelength of a photon depends on the energy transferred by a decelerating electron
    • The electrons don't all undergo the same deceleration when they strike the target
    • This leads to a distribution of energies, hence, a range, or continuous spectrum, of wavelengths is observed

Part (b)

Step 1: Identify the significance of the intensity 

    • The intensity of the graph signifies the proportion of photons produced with a specific energy, or wavelength
    • The higher the intensity, the more photons of a particular wavelength are produced
    • In other words, the total intensity is the sum of all the photons with a particular wavelength

Step 2: Explain the shape of the graph

    • When a single electron collides with the metal target, a single photon is produced
    • Most electrons only give up part of their energy, and hence there are more X-rays produced at wavelengths higher than the minimum (or energies lower than the maximum)
    • At short wavelengths, there is a steeper gradient because only a few electrons transfer all, or most of, their energy

Part (c)

Step 1: Identify the relationship between minimum wavelength and maximum energy

    • The minimum wavelength of an X-ray is equal to

lambda subscript m i n end subscript equals fraction numerator h c over denominator E subscript m a x end subscript end fraction

    • The equation shows the maximum energy of the electron corresponds to the minimum wavelength, they are inversely proportional 

lambda subscript m i n end subscript proportional to 1 over E subscript m a x end subscript

    • Therefore, the higher the energy of the electron, the shorter the wavelength of the X-ray produced

Step 2: Explain the presence of the cut-off point

    • The accelerating voltage determines the kinetic energy which the electrons have before striking the target
    • The value of this accelerating voltage, therefore, determines the value of the maximum energy
    • This corresponds to the minimum, or cut-off, wavelength

You've read 0 of your 0 free revision notes

Get unlimited access

to absolutely everything:

  • Downloadable PDFs
  • Unlimited Revision Notes
  • Topic Questions
  • Past Papers
  • Model Answers
  • Videos (Maths and Science)

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Katie M

Author: Katie M

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.