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.

----

I know the relation between d and h+k+l
but what does it mean the collimated beam stuff?

A collimated beam refers to a beam of light or radiation that is parallel and does not diverge or converge. In the context of X-ray diffraction, a collimated beam is achieved by passing the X-rays through a collimator, which is a device that restricts the radiation to a narrow, parallel path.

By exposing the {111} planes of platinum to a collimated beam of X-rays, we can observe the diffraction pattern produced. From this diffraction pattern, we can determine the angle at which the monochromatic CrKα radiation is obtained.

To calculate this angle, we can use Bragg's law, which states:

nλ = 2d sin(θ)

In this equation, n is an integer representing the order of the diffraction peak, λ is the wavelength of the radiation, d is the interplanar spacing of the crystal planes, and θ is the angle of diffraction.

In the case of {111} planes of platinum, we need to calculate the interplanar spacing (d) and the corresponding diffraction angle (θ).

The interplanar spacing (d) for a set of crystal planes can be determined using Miller indices (hkl) and the lattice parameter (a). In this case, the Miller indices are {111}. For a face-centered cubic (FCC) crystal system like platinum, the relationship between the Miller indices and the interplanar spacing is:

d = a / √(h^2 + k^2 + l^2)

Since the crystal structure of platinum is FCC, we need to find the lattice parameter (a) of platinum. The lattice parameter can be determined experimentally or obtained from reliable sources such as crystallographic databases.

Once you have the value of the lattice parameter (a), substitute it into the equation to calculate the interplanar spacing (d) for {111} planes.

Then, to find the diffraction angle (θ), rearrange Bragg's law:

θ = arcsin(nλ / (2d))

Substitute the values of n (in this case, 1 for the first-order peak) and the wavelength (λ) of the monochromatic CrKα radiation to calculate the diffraction angle (θ) for platinum {111} planes when exposed to the collimated X-ray beam.

Remember to convert the angle from radians to degrees to obtain the final answer.