Message-ID: <1054819421.158215.1606607354242.JavaMail.confluence1@it-confluence-01.desy.de> Subject: Exported From Confluence MIME-Version: 1.0 Content-Type: multipart/related; boundary="----=_Part_158214_1822862245.1606607354241" ------=_Part_158214_1822862245.1606607354241 Content-Type: text/html; charset=UTF-8 Content-Transfer-Encoding: quoted-printable Content-Location: file:///C:/exported.html HowTo: Alignment of in =E2=80=93 and off-plane reflexes

# HowTo: Alignment of in =E2=80=93 and off-plane reflexes

Alignment of in =E2=80=93 and off-plane reflexes in 6-circle dif= ractometer setup at P10

In case of a symmetric reflection, when diffraction planes of a= reflection are parallel to the surface of a crystal, finding the = reflection is quite easy. Let us consider the vertical scattering geometry = when the =E2=80=98omega=E2=80=99 circle is used for angular scanning of the= sample crystal and the =E2=80=98delta=E2=80=99 circle is used for position= ing the detector.

1)        First step is to align the = crystal in the direct beam. For that both =E2=80=98omega=E2=80=99 and =E2= =80=98delta=E2=80=99 should be set at 0. The detector slits should be open = to a size of 5x5 mm^2 and the detector arm (=E2=80=98delta=E2=80=99) has to= be scanned (without sample in the beam) to define the detector zero-positi= on d0. For further finding zero-=E2=80=98omega=E2=80=99 position= , the vertical size of detector slit has to be reduced down to a size of 1 = mm.

2)        The sample (here we suppose= Si crystal slab with <111> orientation of the surface) should be set= by =E2=80=98cryoz=E2=80=99 (vertical sample translation) to a half-intensi= ty position; then the rocking scan by =E2=80=98omega=E2=80=99 should be don= e and =E2=80=98omega=E2=80=99 position has to be corrected to the peak inte= nsity position. Above steps (starting from setting a sample to half-intensi= ty position) should be repeated several times with decreasing stepwidth unt= il the exact zero-=E2=80=98omega=E2=80=99 position (w0) is defin= ed.

3)        Next is to set the =E2=80= =98omega=E2=80=99-circle to the position w0+qb and th= e =E2=80=98delta=E2=80=99-circle to the position d0+2qb , where qb is the Bragg angle for the aimed reflection [examp= le Si(111)] at the working X-ray energy.

4)        In order to find the reflec= tion, one has to scan =E2=80=98omega=E2=80=99 in a reasonable angular range= with a step matching the expected width of the rocking curve; normally the= reflection is found after scanning =E2=80=98omega=E2=80=99 within =C2=B10.= 1=C2=B0 and a step of 0.001=C2=B0.

In case of an asymmetric reflection, when diffraction planes of= a reflection are not parallel to the surface of a crysta= l, finding the reflection is a bit more elaborate than for symmetric case. = The first step is to find out whether the desired reflection is possible fo= r the actual surface orientation and applied photon energy. As it is illust= rated in Figure 1, the reflection will be possible to measure (in reflectio= n geometry) if its Bragg angle is larger than the angle j between the diffr= action planes and the surface (otherwise the reflection can be measured onl= y in Laue (transmission) geometry). Figure 1: Example of lattice planes.

For a cubic lattice the angle j between two planes having Miller indexes= h1k1l1 and h2k2l2 is defined as:

cos(j)=3D (h1*h2 + k1*k2 + l= 1*l2 )/(sqrt(h1**2 + k1**2 + l<= sub>1**2)*sqrt(h2**2 + k2**2+ l2**2)= )

For example, for the Si(311) reflection and a <111> crystal surfac= e orientation we have j=3D29.5=C2=B0. For X-ray energy of 8 keV the Bragg a= ngle of Si(311) reflection is qb =3D 28.2=C2=B0, which means tha= t this reflection is not accessible in Bragg (reflection) geometry at this = X-ray energy.

After the measurable reflection is selected, the steps 1) and 2) should = be applied. As before, in step 3) the detector arm has to be places at d0+2qb. But the =E2=80=98omega=E2=80=99 circle should be = positioned either at qb-j or at qb+j (see Figure 1). = The case of incident angle equal to qb-j is favourable sin= ce the intrinsic width is becoming larger due to asymmetry factor; and henc= e the reflection will be easier to find. After the =E2=80=98delta=E2=80=99 = and =E2=80=98omega=E2=80=99 circles are brought to their positions, one has= to start azimuthal scans of the sample using the =E2=80=98phi=E2=80=99 cir= cle. Scanning of =E2=80=98phi=E2=80=99 motor should be performed in a wide = angular range of 180=C2=B0 with reasonably small step of 0.01=C2=B0.=   After the reflection is found, one can measure its rocking curve by = scanning =E2=80=98omega=E2=80=99 as in step 4).

## Example: Si(111) & Si(444) reflexes

As an example, the rocking curve of Si(111) and Si(444) at E=3D8keV are = displayed.  ------=_Part_158214_1822862245.1606607354241--