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Modern method of production of X-rays:

It consist of an evacuated glass tube of about pressure 10-5mm of Hg. The cathode consist of tungsten filament (F)
heated by L.T.B. The high voltages is applied between filament and Target (T). During continuous bombardment of the electron
on the target, heat is generated, in order to maintain the temperature of the target the cold water is circulated through the tube.
The filament is placed inside the metal cap to focus the electron into the target.

WORKING: When electric current flowing through the filament by L.T.B, heat is generated in it. And hence electrons are
emitted and accelerated towards the target by high voltage and strike the target with high velocity.
As a result, X-rays are emitted from the target.

X-ray diffraction (Laue experiment) (Wave nature of X-ray)

In 1913, van laue shows that if a beam of X-rays is allowed to pass through the crystal, the diffraction pattern is obtained.
This experiment consist of X-ray tube (S) which emitts continuous range of wavelength. A narrow beam of X-rays from the tube is emitted
by two slits S1 and S2 and is allowed to pass through ZnS (Zinc Sulphite) crystal and transmitted beam of X-rays
is received on photographic plate. After explosure of several hours a laue pattern is obtained on the photographic plate. The photograph
consist of central spot which arises due to direct beam of X-rays.

The central beam is surrounded by fainter spot in different pattern. These spots are called laue spot. These indicates that incident beam of
X-rays has been difracted from various plane. These spots are arranged according to different geometrical pattern.

Result: This experiment shows that:
1. X-rays are electromagnetic wave.
2. The atom of the crystal are arranged in three dimensional lattice.

Bragg's Law:

When beam of monochromatic X-rays is incident upon the atoms in the crystallatice, each atoms act as a source of scattering radiation of same wavelength.
The crystal as the series of parallel reflecting planes. The intensity of reflected beam of certain angle will be maximum when path difference
between two reflected waves from two different planes is an integral multiple of wavelength.
Let, PQ,RS and UV represent three atomic planes of a crystallatice constant 'd'. Let, a beam of monochromatic X-rays of wavelength is incident on
the crystal making an glancing angle. Suppose two waves AB and CD are incident on planes PQ and RS reflected along BE and DF.

Now in order to find the path difference, draw BM perpendicular to CD and BN perpendicular to DF. Therefore the path difference between these
two waves is given by,
Path difference= MD + DN <--------- (i)
from ΔBMD
sinθ = MD/BD
MD= BDsinθ
MD= dsinθ

from ΔBND
sinθ= DN/BD
DN= BDsinθ
DN= dsinθ

So equation (i) becomes,
path difference= MD+DN
= dsinθ + dsinθ
= 2dsinθ

For maximum intensity, path difference = nλ
where n is the integer and its value is 1,2,3,......., ∞

* For first order diffraction
n= 1 and θ = θ
So, 2dsinθ = λ

* For maximum order diffraction
θ = 90o
i.e. 2dsin90o = nλ
i.e. n = 2d/λ

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