Modern Physics Concept Page - 7

Example

Problem on variation of energy and wavelength of X-rays

Example:
The voltage applied to an X-ray bulb increases to 1.5 times the initial voltage. The short wavelength limit of continuous X-ray spectrum shifts by 25 pm. Then find the original wavelength.Solution:
λ=hceV$\lambda ={\displaystyle \frac{hc}{eV}}$
λ−25=2hc3eV$\lambda -25={\displaystyle \frac{2hc}{3eV}}$
λλ−25=32${\displaystyle \frac{\lambda }{\lambda -25}}={\displaystyle \frac{3}{2}}$
2λ= 3λ−75$2\lambda =3\lambda -75$
λ=75 pm$\lambda =75pm$

Definition

Moseley's law and Bohr model

Moseley's Law:
ν√=a(Z−b)$\sqrt{\nu }=a(Z-b)$
where a and b are constants
and ν$\nu$ is frequency of X-rays
According to Bohr's model, the energy released during the transition from n=2$n=2$ to n=1$n=1$ is given by
ΔE=hν=Rhc(Z−b)2(112−122)$\mathrm{\Delta }E=h\nu =Rhc{(Z-b)}^2({\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{2^2}})$
so that ν√=3Rc4−−−−√(Z−b)$\sqrt{\nu }=\sqrt{{\displaystyle \frac{3Rc}{4}}}(Z-b)$

Definition

Laue Experiment

X-rays from the X-ray tube is passing through the two slits of two tubes. The beam is now allowed to pass through the ZnS$ZnS$ crystal. The emergent rays are made to fall on to a photographic plate. The diffraction pattern so obtained consist of a central spot at the centre and series of spots arranged in a definite pattern about the centre. The central spot is due to the direct beam, whereas the regularly arranged spots are due to the diffraction pattern from the atoms of the various crystal planes. This spots are called as Laue spots.
Two important facts, they are
 1. X-ray are electromagnetic waves of externally short wave length.
2. the atoms in a crystal are arranged in a regular three dimensional lattice.

Definition

Compton wavelength

The Compton wavelength of a particle is equivalent to the wavelength of a photon whose energy is the same as the rest-mass energy of the particle. The standard Compton wavelength, of a particle is given by:

λ=hmec=0.00243 nm$\lambda ={\displaystyle \frac{h}{m_{e}c}}=0.00243nm$

Formula

Change in wavelength due to Compton effect

Δλ=hmec(1−cosθ)$\mathrm{\Delta }\lambda ={\displaystyle \frac{h}{m_{e}c}}(1-cos\theta )$

Example

Bragg's law and its application

Suppose, and X-ray bean is incident on a solid, making an angle θ$\theta$ with the planes of the atoms. These X-rays are diffracted by different atoms and the diffracted rays interfere. In certain directions, the interference is constructive and we obtain strong reflected X-rays. The analysis shows that there will be a strong reflected X-ray beam only if
2dsinθ=nλ$2d\sin \theta =n\lambda$
where n$n$ is an integer. this equation is known as Bragg's law.
Example: X-rays are known to be electromagnetic radiation with wavelength of the order of 1A∘$1A^{\circ }$. They are produced when accelerated electrons strike target inside an evacuated tube. It is well known that when an electron is accelerated through a potential difference of V it acquires energy eV. If all this energy is used in producing one quantum of X-radiation, then hf=eV$hf=eV$. It is likely that the electron may have lost some of its acquired energy before producing the quantum of radiation. The f gives the maximum possible frequency of X-radiation emitted and that corresponds to the short wavelength limit of the emitted spectrum. In general, therefore, X-ray spectra consist of a continuous spectrum upon which is superposed a line spectrum that is characteristic of the element used as target. For some time after the discovery of X-rays, there was considerable speculation about the nature of X-rays. Max Von Laue found, in 1912 that if X-rays are passed through a crystal they get diffracted. As Laue
patterns are difficult to interpret, Bragg worked out a simple equation that predicts the conditions under which diffracted X-rays beams from a crystal are possible. In its simplest form, Bragg's Law is given by λ=2dsinθ$\lambda =2d\sin \theta$ where d is the perpendicular distance between the planes contining atoms and θ$\theta$ is the glancing angle at which the X-rays fall on the crystal. λ$\lambda$ is known, the distance may be found from experimental measurements. This is the basis for the field of X-ray crystallography in which the structure of crystals is determined by using X-rays. Find the photon is the order of energy associated with an x-ray.

Solution:
Using the equations
E=hcλ$E={\displaystyle \frac{hc}{\lambda }}$
E=eV$E=eV$ and equating we get:
V=hceλ=6.63×10−34×3×10810−10×1.6×10−19=12,413V≈10KeV(order)$V={\displaystyle \frac{hc}{e\lambda }}={\displaystyle \frac{6.63\times {10}^{-34}\times 3\times {10}^8}{{10}^{-10}\times 1.6\times {10}^{-19}}}=12,413V\approx 10KeV(order)$

Definition

The working and construction of Bragg's X-ray spectrometer

Braggs spectrometer used to determine the wavelength of X rays is shown in Fig. Braggs spectrometer is similar in construction to an ordinary optical spectrometer.
Xrays from an X-ray tube are made to pass through two fine slits S1$S1$ and S2$S2$ which collimate it into a fine pencil. This fine X-ray beam is then made to fall upon  the  crystal  C$C$ (usually sodium chlo- ride crystal) mounted on       the  spectrometer table. This table is capable of rotation about a vertical axis and  its rotation can be read on a circular graduated scale S. The reflected  beam after passing through the  slits  S3$S3$  and  S4$S4$ enters the ionization chamber. The X-rays entering the ionization chamber ionize the gas which causes a current to flow between the electrodes and the current can be measured by galvanometer G$G$. The ionization current is a measure of the intensity of X-rays reflected by the crystal.
The ionization current is measured for different values of glancing angle

$.Agraphisdrawnbetweentheglancingangle$$\$.Agraphisdrawnbetweentheglancingangle$$

$ and ionization current.
For certain values of glancing angle,the ionization current increases abruptly. The first peak corresponds to first order, the second peak to second order and so on. From the graph, the glancing angles for different orders of reflection can be measured. Knowing the angle and the spacing d for the crystal, wavelength of Xrays can be determined.

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