Awainaideannagi

2022-07-20

Suppose you want to produce a diffraction pattern with X rays whose wavelength is 0.025 nm. If you use a diffraction grating, what separution between lines is needed to genorate a pattern with tne first maximum at an angle of ${14}^{\circ}$ ?

Alden Holder

Beginner2022-07-21Added 15 answers

X-ray diffraction :

In the case of X-ray diffraction crystal is used instead of the diffraction grating. The crystal consists of an array of equally spaced atoms that act as a grating. The incident light is used in such a way that its wavelength is comparable with an interatomic plane of the crystal.

When such a x-ray incident on the crystal at a particular angle then it reflects at different planes of the atom. The reflected rays for a particular set of planes interfere with each other. If these reflected rays interfere constructively it gives x-ray signal i.e. sharp peak.

The condition to get this pattern/peak is;

$2d\mathrm{sin}(\theta )=n\lambda $

where, d= distance between interatomic planes

Here $\theta $ is the angle between normal to the plane and incident (and reflected) X - rays

Therefore, $2\mathrm{sin}(\theta )$ insted of only $\mathrm{sin}(\theta )$

In the given problem, the conditions given are the same as that of x-ray diffraction.

Formula:

$2d\mathrm{sin}(\theta )=n\lambda $

Given:

$\theta ={14}^{\circ}\phantom{\rule{0ex}{0ex}}n=1\phantom{\rule{0ex}{0ex}}\lambda =0.025\text{nm}\phantom{\rule{0ex}{0ex}}d=?\phantom{\rule{0ex}{0ex}}d=\frac{n\lambda}{2\mathrm{sin}(\theta )}\phantom{\rule{0ex}{0ex}}d=\frac{1\times 0.025\text{nm}}{2\times \mathrm{sin}(14)}\phantom{\rule{0ex}{0ex}}d=\frac{0.025}{0.434}\phantom{\rule{0ex}{0ex}}d=0.0576\text{nm}$

Answer: The required separation between lines is $d=0.0576\text{nm}$

In the case of X-ray diffraction crystal is used instead of the diffraction grating. The crystal consists of an array of equally spaced atoms that act as a grating. The incident light is used in such a way that its wavelength is comparable with an interatomic plane of the crystal.

When such a x-ray incident on the crystal at a particular angle then it reflects at different planes of the atom. The reflected rays for a particular set of planes interfere with each other. If these reflected rays interfere constructively it gives x-ray signal i.e. sharp peak.

The condition to get this pattern/peak is;

$2d\mathrm{sin}(\theta )=n\lambda $

where, d= distance between interatomic planes

Here $\theta $ is the angle between normal to the plane and incident (and reflected) X - rays

Therefore, $2\mathrm{sin}(\theta )$ insted of only $\mathrm{sin}(\theta )$

In the given problem, the conditions given are the same as that of x-ray diffraction.

Formula:

$2d\mathrm{sin}(\theta )=n\lambda $

Given:

$\theta ={14}^{\circ}\phantom{\rule{0ex}{0ex}}n=1\phantom{\rule{0ex}{0ex}}\lambda =0.025\text{nm}\phantom{\rule{0ex}{0ex}}d=?\phantom{\rule{0ex}{0ex}}d=\frac{n\lambda}{2\mathrm{sin}(\theta )}\phantom{\rule{0ex}{0ex}}d=\frac{1\times 0.025\text{nm}}{2\times \mathrm{sin}(14)}\phantom{\rule{0ex}{0ex}}d=\frac{0.025}{0.434}\phantom{\rule{0ex}{0ex}}d=0.0576\text{nm}$

Answer: The required separation between lines is $d=0.0576\text{nm}$

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