Bragg's Law- Solved Examples

William Lawrence Bragg and his father, William Henry Bragg, are physicists who are given credit for developing Bragg’s Law as a foundational principle of atomic and nuclear structure studies. The law explains the way in which crystals diffract X-rays, in the materials through the interference due to them (X-rays.) They are said to be scattered to them by crystals if they encounter such substances. What is to be done for constructive interference to take place is to make sure that the path difference between these scattered rays is a whole number multiple.

Bragg's Law is an important concept in the field of atom and atomic nuclei studies and therefore students appearing in Class 12, NEET, and JEE Main exams should be familiar with it. It explains how X-rays are bent when they interact with atomic planes of crystals thereby disclosing the arrangement of these crystals. Over the last ten years of the JEE MAIN exam (from 2013 to 2023), a total of one question has been asked on this concept.

Bragg's law

X-ray is used in measuring the interplanar spacing 'd' and several information about the structure of the solid can be obtained. This phenomenon can be understood by Bragg's law.

As we have learned Bragg's law already in the Dual Nature Of Matter And Radiation, here also 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 -

2dsin⁡θ=nλ

where n is an integer. For monochromatic X -rays, λ is fixed and there are some specific angles θ1,θ2,θ3,… etc. corresponding to n=1,2,3,… etc. in the equation given above. Thus, if the X -rays are incident at one of these angles, they are reflected; otherwise, they are absorbed.

Derivation of Bragg's Law

Consider the following figure of beams. The phases of the beams coincide when the incident angle equals the reflecting angle. The incident beams remain parallel to each other until they reach point z. At this point, they strike the surface and travel upwards. At point B, the second beam scatters. The distance travelled by the second beam is $A B+B C$. This additional distance is known as the integral multiple of the wavelength.

$$
\mathrm{n} \lambda=\mathrm{AB}+\mathrm{BC}
$$
We also know that $A B=B C$

$$
\mathrm{n} \lambda=2 \mathrm{AB} \text { (equation 1) }
$$

$d$ is the hypotenuse of the right triangle $A b z . A b$ is the opposite of the angle $\theta$.

$$
A B=d \sin \theta \text { (equation 2) }
$$
Substituting equation 2 in equation 1 , we get:

$$
\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta
$$
This is the final expression of Bragg's law.

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Solved Examples Based on Bragg's Law

Example 1: Bragg's law is given by the equation
1) nλ=2θsin⁡θ
2) nλ=2dsin⁡θ
3) 2nλ=dsin⁡θ
4) nθ2=d2sin⁡θ

Solution:

Bragg's law

2dsin⁡θ=nλ

where n is an integer. For monochromatic X-rays, λ is fixed and there are some specific angles θ1,θ2,θ3,…, etc. corresponding to n=1,2,3,…, etc. in the equation given above. Thus, if the X-rays are incident at one of these angles, they are reflected; otherwise, they are absorbed.

Hence, the answer is the option (2).

Example 2: A beam of electrons of energy E scatters from a target having atomic spacing of 1A. The first maximum intensity occurs at θ=60∘. Then E( in eV) is

(Plank constant h=6.64×10−34J s1eV=1.6×10−19 J electron mass m=9.1×10−31 kg)

1) 50.47

2) 100.94

3) 40.47

4) 80

Solution:

2dsin⁡θ=λ=h2mE2×10−10×32=6.6×10−342mEE=12×6.642×10−489.1×10−31×3×1.6×10−19=50.47

Example 3: In X-ray diffraction, first maxima occur at θ=30∘ then the wavelength of X-ray are

(d=1A)

1) 1 A∘
2) 2A∘
3) 3.A∘
4) 0.5A∘

Solution:

Bragg's law

2dsin⁡Θ=nλ
(condition of constructive maxima )
d= distance between parallel lines
λ= wavelength
Θ= angle between light \& plane
Here in X -ray diffraction n=1
2dsin⁡θ=λ
λ=2×1A∘×sin⁡30∘=1A∘

Hence, the answer is the option (1)

Summary

X-rays are scattered by atoms in a crystal lattice and the way they are scattered helps to determine the structure of a crystal as well as their positions in this lattice. If X-rays meet a crystal, according to this law, they turn back from layers of atoms which are in it. And if a beam from a first-order reflection is made according to Bragg law, there comes constructive interference which gives a signal that can be detected.