Question

X-rays of wavelength 0.103 nm reflects off a crystal and a second-order maximum is recorded at a Bragg angle of 25.5°. What is the spacing between the scattering planes in this crystal?

Answer 1

The spacing between the scattering planes in the crystal is approximately 0.186 nm.

Step-by-step explanation:

In X-ray diffraction, the Bragg equation relates the wavelength of X-rays, the angle of incidence, and the spacing between the scattering planes in the crystal. The Bragg equation is given by:

nλ = 2d sin(θ)

where n is the order of the maximum, λ is the wavelength of the X-rays, d is the spacing between the scattering planes, and θ is the Bragg angle.

In this case, the second-order maximum is observed at a Bragg angle of 25.5° and the wavelength of the X-rays is 0.103 nm. To find the spacing between the scattering planes, we need to solve for d:

d = nλ / (2sin(θ))

Substituting the values, we get:

d = (2 × 0.103 nm) / (2sin(25.5°))

Simplifying, we get:

d ≈ 0.186 nm

Therefore, the spacing between the scattering planes in this crystal is approximately 0.186 nm.