Question

An x-ray beam with wavelength 0.170 nm is directed at a crystal. As the angle of incidence increases, you observe the first strong interference maximum at an angle 62.5 ∘. What is the spacing d between the planes of the crystal?

Answer 1

The distance between the scattering planes in the crystal is d = 0.95 A°

Explanation:

The Bragg's equation is given by

                                      2d sinθ = nλ

where,

d is the distance between the scattering planes in the crystal.

θ is the angle of diffraction.

n is the order of diffraction.

λ is the wavelength of X rays.

Given λ = 0.17 nm = 1.7 A°, angle = 62.5

                          2 $\times$ d $\times$ sin(62.5)    = 1 $\times$ 1.7 A°  

                                                    d = 0.95 A°  

The distance between the scattering planes in the crystal is d = 0.95 A°

Answer 2

The spacing (d) between the planes of the crystal is 0.096nm.

BRAGG'S LAW EQUATION:

• The spacing or distance between the planes of the crystal can be calculated using Bragg's law equation as follows:

2dsinθ = nλ

Where;

1. d = distance between the scattering planes in the crystal
2. θ is the angle of diffraction
3. n is the order of diffraction
4. λ is the wavelength of X rays

d = nλ ÷ 2sinθ

According to this question;

1. n = 1
2. λ = 0.170nm
3. θ = 62.5°

• d = (1 × 0.170) ÷ 2 × sin 62.5°

• d = 0.170 ÷ 1.77

• d = 0.096nm

• Therefore, the spacing (d) between the planes of the crystal is 0.096nm.

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