Contents
Physics Topics can help us understand the behavior of the natural world around us.
What do you Mean by Hard and Soft X-rays?
Roentgen’s discovery: In 1895, German physicst Wilhelm Roentgen observed that when high velocity cathode rays were obstructed by any solid target, an invisible ray came out from that target. The characteristics of this ray that he observed are: The rays
- can produce fluorescence,
- can affect photographic plates,
- can penetrate thin sheets of light materials like paper, glass, wood, aluminium, etc.
- cannot penetrate heavy metals like iron, lead, etc. Rather it casts shadow behind them.
At that time, the nature of this ray was unknown, and hence Roentgen called this ray X-ray.
Nature of X-rays
X-rays are not deflected by electric or magnetic field; from this we can conclude that, X-rays are not streams of charged particles.
So, X-ray is either a stream of high velocity uncharged particles or a kind of wave.
If X-ray is considered to be a kind of wave, it should show properties like interference, diffraction, etc. But in any traditional experimental set up, these properties were not observed, until finally Max von Laue and others after him observed diffraction of X-ray by passing it through three-dimensional crystals.
Thus it was proved that X-ray is a kind of wave. The wavelength of this ray is so small that its diffraction is possible only by crystals of regular intermolecular space (10-8 cm approximately). Later on it has been possible to observe the wave-properties of X-rays like reflection, refraction, interference, and polarisation.
So, it has been observed, X-ray is not a stream of any high velocity uncharged particles, but an electromagnetic wave like visible light. Though compared to the visible light, the wavelength of X-ray is very small, almost \(\frac{1}{1000}\) parts or even less than that of the former. It was not until 1923 when A. H. Compton, by means of his X-ray scattering experiment, established the particle nature of photon, and hence of X-rays.
Frequency and Wavelength of X-ray
According to quantum theory, electromagnetic radiation is a stream of massless and chargeless photons which travels with the speed of light(c = 3 × 108m ᐧ s-1).
Energy of each photon,
E = hf = \(\frac{h c}{\lambda}\) ….. (1)
Here, h = Planck’s constant = 6.63 × 10-34 J ᐧ s;
f = frequency and λ = wavelength of the electromagnetic wave.
When an electron (charge = e) overcomes a potential difference V, energy gained by it = eV. If this energy is spent entirely to eject an X-ray photon, the energy and frequency of the photon become maximum. If the corresponding wavelength be taken as λmin, then
eV = hfmax = h\(\frac{c}{\lambda_{\min }}\)
or, λmin = \(\frac{h c}{e V}\) …. (2)
In SI, e = 1.6 × 10-19C, c = 3 × 108m ᐧ s-1 and h = 6.63 × 10-34 J ᐧ s
If the potential difference between the anode and the cathode be V, putting these values in equation (2) we get,
λmin = \(\frac{1.243 \times 10^{-6}}{V}\)m
= \(\frac{12430}{V}\)Å = \(\frac{1243}{V}\)nm … (3)
For example,
if V = 50 kV = 50000V, then λmin = 0.2486 Å;
if V = 10 kV = 10000V, then λmin = 1.243 Å;
Discussions:
i) When the whole energy of an electron (eV) belonging to cathode rays is utilised to bombard the target to produce a photon, only then the wavelength of the X-ray becomes equal to λmin as given in equation (2). Generally, the
whole energy is not converted into the energy of X-ray photon; and hence the wavelength becomes more than λmin. For this, λmin is called minimum wavelength of X-ray.
ii) Higher the value of the potential difference V, smaller will be the wavelength of X-rays and hence greater will be its penetrating power. From equation (3), we see that, V should be more than 106 volts to produce hard X-rays equivalent to the wavelength 0.01Å.
iii) Wavelength of X-ray is \(\frac{1}{1000}\) part of the wavelength of visible light. Hence, from the equation, E = \(\frac{h c}{\lambda}\), we can say that, the energy of an X-ray photon is about 1000 times greater.
iv) Soft and hard X-rays: The wavelength range of X-rays is from 0.01Å to 10Å. If λ ≈ 10 Å, according to the relation E = \(\frac{h c}{\lambda}\), the energy of X-ray photon becomes 1240 eV or 1.24 keV (approx.); this kind of X-ray has less penetrating power and is called soft X-ray. On the other hand, if λ ≈ 0.01 Å, the energy of X-ray photon becomes 1.24 MeV (approx.). Due to this high energy, the penetrating power of X-ray becomes very high. It is known as hard X-ray.
Uses of X-rays:
Important uses of X-rays in medical science:
- Radiography: X-rays are used for detection of fractured bones and kidney stones.
- Radiotherapy: Hard X-rays are used in radiotherapy to destroy the cancer affected cells.
- Fluoroscopy: Fluoroscopy is an Imaging technique in which X-rays are used to take real-time moving images of the internal structures of a Living body.
Important uses of X-rays in other fields:
- X-rays are used to analyse structures of different crystals.
- In metallurgy X-rays are used to determine the defects iñ metallic castings.
X-Rays and Atomic Number
When a solid target (made of solid copper and tungsten) is bombarded with a stream of high velocity electrons having kinetic energy of some keV or higher X-rays are emitted from the target. Sometimes, this kind of materials absorb X-rays. Analysing the emitted or absorbed X-rays, we come to know many aspects of the atomic structure of these elements.
Consider a molybdenum (Mo) target being bombarded by a stream of electrons having kinetic energy of 35 keV. The spectrum of the wavelengths of X-rays thus produced is shown in Fig. This spectrum is formed due to the superposition of two kinds of spectra:
1. continuous spectrum,
2. characteristic or line spectrum. A continuous X-ray spectrum superimposed with some brighter lines is formed on the photographic plate. The origins of the two spectra are different and explained in the next article.
Continuous X-ray Spectrum
During the discussion of continuous spectrum shown in Fig., the characteristic spectrum consisting two sharp peaks will be overlooked.
Suppose an electron travelling with kinetic energy K0 under goes collision with an atom of molybdenum [Fig.]. Assume that the electron loses ∆K amount of energy by this collision.
This energy is converted into the energy of an X-ray photon. It should be mentioned here that, the atom being much heavier than an electron, the amount of energy transferred to the atom is neglected.
The electron may collide again with another atom after its collision with the first atom. In this case, the electron collides with (K0 – ∆K) amount of energy. The X-ray photon thus emitted generally possesses less energy than that of the previous photon.
In this way, the electron may suffer successive collisions with different atoms until it comes to rest. The photons thus emitted during these collisions having different energies, form a part of the continuous spectrum of X-rays. But in actual practice, a tar-get is hit with innumerable electrons. Hence, the entire X-ray spectrum looks like that shown in Fig.
This kind of spectrum has an Important characteristic. It has a definite cut-off wavelength (say, λmin). Below this cut-off
wavelength, there is no existence of the spectrum. If the electron loses its whole kinetic energy (K0) in the first impact, X-ray of wavelength λmin is emitted. If f be the frequency of the emitted X-ray photon, then
K0 = hf = \(\frac{h c}{\lambda_{\min }}\) or, λmin = \(\frac{h c}{K_0}\) ….. (1)
So, the value of λmin does not depend on the nature of the solid used as the target. So, if copper is used instead of molybdenum as the target material, the continuous X-ray spectrum changes but the value of λmin remains same.
Numerical Examples
Example 1.
If a stream of electrons of kinetic energy 36 keV is bombarded on a molybdenum target, what will be the cut-off wavelength of the emitted X-ray?
Solution:
We know that, λmin = \(\frac{h c}{K_0}\)
where h = 4.14 × 10-15 eV ᐧ s, c = 3 × 108m ᐧ s-1
Given, K0 = 36 keV = 3.6 × 104 eV
∴ λmin = \(\frac{4.14 \times 10^{-15} \times 3 \times 10^8}{3.6 \times 10^4}\)
= 3.45 × 10-11m = 0.0345 nm
Example 2.
What minimum terminal potential difference of a Coolidge tube should be maintained to produce X-ray of wavelength 0.8Åv (h = 6.62 × 10-34 J ᐧ s, e = 1.60 × 10-19C, c = 3 × 108m ᐧ s-1]
Solution:
Energy of X-ray photon,
E = \(\frac{h c}{\lambda}\) [here, A = 0.8Å = 0.8 × 10-10m]
If electrons be accelerated in a X-ray tube by a potential difference V, energy of an electron = eV; if this energy is completely converted into the energy of X-ray photon, the value of V will be the minimum.
So, eVmin = \(\frac{h c}{\lambda}\)
or, Vmin = \(\frac{h c}{e \lambda}\) = \(\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{1.60 \times 10^{-19} \times 0.8 \times 10^{-10}}\) = 15516V