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X Ray Definition

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On 8 November, 1895, X-rays were discovered by a German Physicist named Wilhelm Conrad Röentgen.

X-ray is an electromagnetic radiation with very short wavelength, and very high energy. X rays have a frequency ranging from 30 petahertz to 30 exahertz.

The wavelength of X-rays is shorter than the Ultraviolet rays, and longer than Gamma rays.

So, what is the wavelength of X rays?

X Rays have a wavelength ranging from 10-12 m (picometers) to 10-9 (nanometers).

X-rays have many applications and in this page, we will cover the top  5 uses of X rays with other uses of X Rays in Physics and X-ray characteristics.


X-Ray is also called the Roentgen radiation. It is an electromagnetic radiation with the energy ranging from 124 eV to 124 keV. Where this energy can be written in the form of Joules. However, a wave with this much energy can easily pass from transparent to opaque objects.

X-rays were discovered accidentally by German scientist Roentgen in 1895. In 1901, Roentgen was awarded for his great work in this regard. 

X-rays are highly penetrating electromagnetic radiation and have proved to be a very powerful tool to study the crystal structure, in material research, in the radiography of metals and in the field of medical sciences. 

How are X-Rays Produced?

Roentgen discovered that when X-rays are passed through arms and hands or any other body part, they create crystal clear and detailed images of the inner bones. 

Whenever a doctor performs an X-ray of a patient, an x-ray sensitive film is put on one side of his body and then the x-rays are shot through him. While the skin is transparent, the bones are dense and absorb more x-rays (because of the nature of X-rays to cross the opaque object). This is why the impression of bones is left on the x-ray film while the skin remains invisible in the x-ray. Below figure shows the wavelength of different electromagnetic spectrum including the X-ray wavelength:

[Image will be uploaded soon]


What is the Wavelength of X Rays?

X-rays possess very short wavelengths that vary between 0.03 and 3 nanometers or between 0.02 Å and 100 Å;  however, some x-rays are small like a single atom of an element. 

Properties of X-Rays

X-rays with short wavelengths with high penetrating ability are highly destructive, that’s why they are called hard x-rays. 

The uses of X rays in medicinal purposes possess less penetrating power and have longer wavelengths and are called soft x-rays. X ray waves have a dual nature. We will now discuss the following properties of these radiations:

  • They can cross the materials with more or unchanged.

  • They are not easily refracted.

  • These rays do not get affected by the electromagnetic field.

  • X-rays ionize the surrounding air by discharging electrified bodies.

  • They have very short wavelengths ranging from 0.1 A° to 1 A°. The velocity of X rays are similar to that of visible light, i.e., 186,000 miles/second or 300,000 kilometers/sec.

  • X-rays are produced when a metallic anode is bombarded/broken by very high energy electrons.

  • They can propagate independently, i.e., without any need of a medium.

  • X-rays are diverging rays, i.e., they cannot be focused on a single point

  • These radiations ar invisible, i.e., they cannot be heard or smelt

  • They make a linear path in a free space but they do not carry an electric charge with them.

  • They can cause photoelectric emissions.

  • Intensity of X - rays rely on the number of electrons hitting the target.

  • Continuous spectrum appears because of the retardation of electrons.

What are the Characteristics of X Rays?

Below are the characteristics of X-rays:

1. The characteristics equation for an X-ray is:

              eV = hfm


e = electron charge;

V = accelerating potential

fm = maximum frequency of X radiation

2. The Characteristic Spectrum due to transition of electron from higher to lower state:

 𝜈  = a (z-b)2  (Moseley's Law)


  𝜈 =  wavenumber

b = shielding factor, whose values vary accordingly: 

b = 1 or ka and 7.4 for La

3. Bragg’s Law

2d Sinθ = nλ

Here, θ= angle for a maximum intensity

4. Binding energy or the total mechanical energy

\[\frac{1}{\lambda }=R(z-b)^{2}[\frac{1}{n1^{2}}-\frac{1}{n2^{2}}]\]


\[v=\frac{1}{\lambda }\] = wave number

R = Rydberg’s constant, whose value is 1.0973731568508 × 10 7 per metre.

5. Cut off wavelength or minimum wavelength, where v (in volts) is the potential difference applied to the tube λmin = 12400 / V A°.

5 Uses of X Rays

  • X-rays are used to analyze alloys through the diffraction pattern.

  • X-rays enable doctors to easily detect things such as a bone fracture or sprain in the body.

  • X-rays are used to identify manufacturing defects in tyres.

  • Doctors use X-rays to check flaws in welding joints and insulating materials.

  • Doctors use X-ray to capture the human skeleton defects.

Uses of X Rays in Physics

  • Restoration

  • Medical Science

  • Security

  • Astronomy

  • Industry

FAQ (Frequently Asked Questions)

1. How are X-Rays Produced?

Ans: In an X-ray tube the electrons emitted/radiated from the metallic cathode are accelerated towards the metal target anode by an accelerating voltage of around 50 kV. The high energy electrons (carrying energy in the range of electronVolts) interact with the atoms in the metal target. 

Some electrons come very near to a nucleus in the target and get deviated by the electromagnetic interaction. In this process, which is called bremsstrahlung  or the braking radiation, the electron releases a lot of energy and a photon a.k.a X-day is emitted. 

Also, the energy of the emitted photon can take any value up to a maximum matching the energy of the incident electron.

2. A P.D. of 103 V is Applied to an X-Ray Tube. Determine the Ratio of the De-Broglie Wavelength of the Incident Electrons and the Wavelength of X-rays Produced.

(a) 1/110

(b) 1/100  

(c) 1/104

(d) 1

(Given: e/m = 1.81014 C/kg for an electron)


For incident electron, we have:

λ1 = h/p 

= h/√2Km 

λ1  = h/√ 2meV

The shortest wavelength of X-rays is:

 λ2 = hc/Ve


λ1/λ2 = 1/c√(V/2) (e/m)

Substituting the values, we get;

λ1/λ2 = 1

From our calculation, we conclude that option (d) is correct.