At what angle (in degrees) must {111} planes of platinum (Pt) be exposed to a collimated beam of thee X-rays in order to obtain by diffraction monochromatic CrKα radiation.

To determine the angle at which {111} planes of platinum (Pt) must be exposed to a collimated beam of X-rays to obtain monochromatic CrKα radiation, we'll need to use Bragg's Law and some additional information.

Bragg's Law can be expressed as follows:

nλ = 2d sinθ

Where:
- n is the order of the diffraction peak (usually 1 for primary peaks)
- λ is the wavelength of the X-rays
- d is the interplanar spacing of the crystal lattice
- θ is the angle of incidence of the X-rays on the crystal lattice planes

In this case, we are given that the desired radiation is monochromatic CrKα radiation. The wavelength (λ) of CrKα radiation is approximately 2.291 Å (angstroms) or 0.2291 nm.

Next, we need to find the interplanar spacing (d) of the {111} planes in platinum. The interplanar spacing for any set of crystal lattice planes can be calculated using the formula:

d = a / √(h² + k² + l²)

Where:
- a is the lattice constant
- h, k, and l are the Miller indices which represent the planes of interest.

For platinum, the lattice constant (a) is approximately 3.924 Å or 0.3924 nm.

To find the Miller indices (hkl) for the {111} planes, we need to identify the smallest ratios of integers that satisfy the equation:

(1/h)² + (1/k)² + (1/l)² = 1

For the {111} planes, the Miller indices are {h, k, l} = {1, 1, 1}.

Now, we can calculate the interplanar spacing (d) for the {111} planes in platinum:

d = 0.3924 nm / √(1² + 1² + 1²)
d = 0.3924 nm / √3
d ≈ 0.226 nm

Finally, we can rearrange Bragg's Law to solve for the angle (θ):

θ = arcsin((nλ) / (2d))

In this case, since we want the primary diffraction peak (n = 1), we have:

θ = arcsin((1 * 0.2291 nm) / (2 * 0.226 nm))

Evaluating this expression will give us the angle at which the {111} planes of platinum must be exposed to the X-ray beam to obtain the desired monochromatic CrKα radiation.