Angle of incidence for diffracted beam from (110) planes in simple cubic lattice

Introduction: The diffraction of X-rays from crystals can be used to determine the arrangement of atoms in a crystal lattice. When an X-ray beam is incident on a crystal, it interacts with the atoms in the crystal and produces a diffracted beam. The angle of incidence at which the diffracted beam is produced depends on the wavelength of the X-rays and the spacing between the atoms in the crystal lattice.

Given: Wavelength of X-rays = 5.0 Å, Simple cubic lattice with a = 10 Å, (110) planes

Formula: The Bragg equation relates the angle of incidence (θ) to the wavelength of the X-rays (λ) and the spacing between the atoms in the crystal lattice (d).

nλ = 2d sin θ

Where n is an integer that determines the order of the diffraction.

Solution:

To find the angle of incidence at which X-rays of wavelength 5.0 Å will produce a diffracted beam from the (110) planes in a simple cubic lattice with a = 10 Å, we need to calculate the spacing between the (110) planes.

The (110) planes of a simple cubic lattice are perpendicular to the [110] direction. The distance between adjacent atoms along the [110] direction is given by:

a[110] = a√2 = 10√2 Å

The spacing between the (110) planes is given by the reciprocal of the distance along the [110] direction:

d110 = 1/a[110] = 1/(10√2) Å

d110 = 0.0707 Å

Now we can use the Bragg equation to calculate the angle of incidence at which X-rays of wavelength 5.0 Å will produce a diffracted beam from the (110) planes in a simple cubic lattice with a = 10 Å.

For n = 1:

nλ = 2d sin θ

θ = sin-1(nλ/2d)

θ = sin-1(5.0/2(0.0707))

θ = 20.7°

For n = 2:

nλ = 2d sin θ

θ = sin-1(nλ/2d)

θ = sin-1(5.0/2(0.0707)) * 2

θ = 45.0°

Therefore, the correct answers are options C and D.