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uhoh
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I happen to have a slide about something similar, so I'll adjust it and paste below.

If the red and green lines are the (acute) 2D lattice vectors for the 2D hexagonal lattice on the fcc(111) face, we can call those directions [1, 0] and [0, 1]. Then "orange" direction could be called [1, 1]. As far as I can tell from the two images below, in the 3D lattice those directions could be called either [1, -1, 0] & [-1, 0 1] or [1, -1, 0] & [1, 0 -1].

Either way, add them together to get the orange arrow and it's [0, -1, 1] or [2, -1, -1] and the rules for allowed diffraction are all odd or all even and the orange line does not satisfy that

Before the intuitive answer, let's look at the selection rules for diffraction We can trust them while we learn to understand them.

Bravais lattices           Allowed reflections           Forbidden reflections

Simple cubic               Any h, k, ℓ                   None

Body-centered cubic (BCC)  h + k + ℓ = even              h + k + ℓ = odd

Face-centered cubic (FCC)  h, k, ℓ all odd or all even   h, k, ℓ mixed odd and even

Diamond FCC                All odd, or all even with     h, k, ℓ mixed odd and even 
                           h + k + ℓ = 4n                or all even with h + k + ℓ ≠ 4n

Triangular lattice         ℓ even, h + 2k ≠ 3n           h + 2k = 3n for odd ℓ

For the orange-arrow case in your question let's look again:

OP's diagram with bigger arrows

Now examine my old slide:

For your plane wave with a projected period on the surface equal to the [1, 1] direction, we can see that it passes TWO rows of equally spaced atoms. The reason that the selection rule says this diffraction is forbidden is that diffraction from the alternating rows of atoms cancel each other; they are 180 degrees out of phase.

my old slide


Trying to get the correct [hkl] notation for directions of the hexagonal lattice on the fcc(111) surface. These seem to :

from Diffusive Atomistic Dynamics of Edge Dislocations in Two Dimensions from Reconstruction of steps on the Cu(111) surface induced by sulfur

Sources left: Diffusive Atomistic Dynamics of Edge Dislocations in Two Dimensions, right: Reconstruction of steps on the Cu(111) surface induced by sulfur

I happen to have a slide about something similar, so I'll adjust it and paste below.

If the red and green lines are the (acute) 2D lattice vectors for the 2D hexagonal lattice on the fcc(111) face, we can call those directions [1, 0] and [0, 1]. Then "orange" direction could be called [1, 1]. As far as I can tell from the two images below, in the 3D lattice those directions could be called either [1, -1, 0] & [-1, 0 1] or [1, -1, 0] & [1, 0 -1].

Either way, add them together to get the orange arrow and it's [0, -1, 1] or [2, -1, -1] and the rules for allowed diffraction are all odd or all even and the orange line does not satisfy that

Before the intuitive answer, let's look at the selection rules for diffraction We can trust them while we learn to understand them.

Bravais lattices           Allowed reflections           Forbidden reflections

Simple cubic               Any h, k, ℓ                   None

Body-centered cubic (BCC)  h + k + ℓ = even              h + k + ℓ = odd

Face-centered cubic (FCC)  h, k, ℓ all odd or all even   h, k, ℓ mixed odd and even

Diamond FCC                All odd, or all even with     h, k, ℓ mixed odd and even 
                           h + k + ℓ = 4n                or all even with h + k + ℓ ≠ 4n

Triangular lattice         ℓ even, h + 2k ≠ 3n           h + 2k = 3n for odd ℓ

For the orange-arrow case in your question let's look again:

OP's diagram with bigger arrows

Now examine my old slide:

For your plane wave with a projected period on the surface equal to the [1, 1] direction, we can see that it passes TWO rows of equally spaced atoms. The reason that the selection rule says this diffraction is forbidden is that diffraction from the alternating rows of atoms cancel each other; they are 180 degrees out of phase.

my old slide

I happen to have a slide about something similar, so I'll adjust it and paste below.

If the red and green lines are the (acute) 2D lattice vectors for the 2D hexagonal lattice on the fcc(111) face, we can call those directions [1, 0] and [0, 1]. Then "orange" direction could be called [1, 1]. As far as I can tell from the two images below, in the 3D lattice those directions could be called either [1, -1, 0] & [-1, 0 1] or [1, -1, 0] & [1, 0 -1].

Either way, add them together to get the orange arrow and it's [0, -1, 1] or [2, -1, -1] and the rules for allowed diffraction are all odd or all even and the orange line does not satisfy that

Before the intuitive answer, let's look at the selection rules for diffraction We can trust them while we learn to understand them.

Bravais lattices           Allowed reflections           Forbidden reflections

Simple cubic               Any h, k, ℓ                   None

Body-centered cubic (BCC)  h + k + ℓ = even              h + k + ℓ = odd

Face-centered cubic (FCC)  h, k, ℓ all odd or all even   h, k, ℓ mixed odd and even

Diamond FCC                All odd, or all even with     h, k, ℓ mixed odd and even 
                           h + k + ℓ = 4n                or all even with h + k + ℓ ≠ 4n

Triangular lattice         ℓ even, h + 2k ≠ 3n           h + 2k = 3n for odd ℓ

For the orange-arrow case in your question let's look again:

OP's diagram with bigger arrows

Now examine my old slide:

For your plane wave with a projected period on the surface equal to the [1, 1] direction, we can see that it passes TWO rows of equally spaced atoms. The reason that the selection rule says this diffraction is forbidden is that diffraction from the alternating rows of atoms cancel each other; they are 180 degrees out of phase.

my old slide


Trying to get the correct [hkl] notation for directions of the hexagonal lattice on the fcc(111) surface. These seem to :

from Diffusive Atomistic Dynamics of Edge Dislocations in Two Dimensions from Reconstruction of steps on the Cu(111) surface induced by sulfur

Sources left: Diffusive Atomistic Dynamics of Edge Dislocations in Two Dimensions, right: Reconstruction of steps on the Cu(111) surface induced by sulfur

Source Link
uhoh
  • 6.1k
  • 4
  • 31
  • 88

I happen to have a slide about something similar, so I'll adjust it and paste below.

If the red and green lines are the (acute) 2D lattice vectors for the 2D hexagonal lattice on the fcc(111) face, we can call those directions [1, 0] and [0, 1]. Then "orange" direction could be called [1, 1]. As far as I can tell from the two images below, in the 3D lattice those directions could be called either [1, -1, 0] & [-1, 0 1] or [1, -1, 0] & [1, 0 -1].

Either way, add them together to get the orange arrow and it's [0, -1, 1] or [2, -1, -1] and the rules for allowed diffraction are all odd or all even and the orange line does not satisfy that

Before the intuitive answer, let's look at the selection rules for diffraction We can trust them while we learn to understand them.

Bravais lattices           Allowed reflections           Forbidden reflections

Simple cubic               Any h, k, ℓ                   None

Body-centered cubic (BCC)  h + k + ℓ = even              h + k + ℓ = odd

Face-centered cubic (FCC)  h, k, ℓ all odd or all even   h, k, ℓ mixed odd and even

Diamond FCC                All odd, or all even with     h, k, ℓ mixed odd and even 
                           h + k + ℓ = 4n                or all even with h + k + ℓ ≠ 4n

Triangular lattice         ℓ even, h + 2k ≠ 3n           h + 2k = 3n for odd ℓ

For the orange-arrow case in your question let's look again:

OP's diagram with bigger arrows

Now examine my old slide:

For your plane wave with a projected period on the surface equal to the [1, 1] direction, we can see that it passes TWO rows of equally spaced atoms. The reason that the selection rule says this diffraction is forbidden is that diffraction from the alternating rows of atoms cancel each other; they are 180 degrees out of phase.

my old slide