1. mass absorption coefficient for lead, m/r = 230 cm²/g

g = 1.54Å

r = 11.3g

I(g) = 10

Io(g) = 10⁸

Equation for X-Ray absorption I(g)/I_o(g) = exp{-(m/r)_grx}

This rearranges to x = ln (I(g)/I_o(g)) / --(m/r)_grr

Substituting the figures into the equation gives

x = 0.00620cm

The extra thickness is used in common practise to reduce the intensity of the CuKb radiation and maintain a good intensity of the CuKa.

2. a) Using a laboratory diffractometer, the use of deep sample holders (if there is plenty of material) or a sample holder that doesn't detract from the intensity of the radiation. The use of Soller slits will prevent angular divergence, keeping the x-rays within the q/2q plane. A beam catcher will also help to minimise any interference caused by non-monochromatic x-rays. The use of a silicon crystal sample holder, leading to a negligible background scatter will aslo help to get a better diffraction pattern.

b) The use of a 'well focussed 'transmission geometry set-up is a good way of collecting data on organic materials with low absorption. This setup utilises a Debye-Scherrer camera to collect data over 360o. this lead to good lattice parameters from very high angle reflection. The set-up uses an incident beam collimator that reduces air scatter at the penalty of focussing the radiation. Curved silicon monochromators can be used to focus the divergent beam and lead to a narrower peaks

3. Constant parameters

g₁ = 0.7Å

g₂ = 0.65Å

n = 1

d = 3.136Å

Equation 1

q₁ = sin^-1 ((ng₁)/2d)

q₁ = sin^-1 (0.7/(2x3.316))

q₁ = 6.408^o

Equation 2

q₂ = sin^-1 ((ng₂)/2d)

q₂ = sin-1 (0.65/(2x3.316))

q₂ = 5.949^o

Difference in angles, Dq = q₁ - q₂

Dq = 0.459^o

 

 

 

 

 

 

4. Constant parameters

E₁ = 20 KeV

E₂ = 100KeV

n = 1

q = 3.5^o

Equation 1

E1 = 12.398/g

g = 12.398/E1 but ng = 2dsinq \ d = (12.398/E1)/2sinq

Hence, d = (12.398/20)/2sin3.5

d = 5.074Å

Equation 2

E2 = 12.398/g g = 12.398/E2 but ng = 2dsinq \ d = (12.398/E2)/2sinq

Hence, d = (12.398/100)/2sin3.5

d = 1.015Å

d-spacing ranging between 1.015Å and 5.074Å can be measures by the detector.

5. I) –x, y, -z.

This is a 2-fold rotation axis parallel to the y direction.

Y will not change, and –x and –z indicate a 180o rotation in those axes. The origin is (0,0,0)

II) ½+x, ½+y, ½+z

This is a point of inversion through (-¼, -¼, -¼)

Considering three points -T_x, -T_y, and -T_z as the centre of origin. The distance to the points, x,y,z. By considering the x-coordinate, the difference between x and the starting position –T_x will be (x--T_x) which is x+ T_x. Once a symmetry operation has taken place, the distance to the new position will similarly be x+ T_x. Therefore, the x co-ordinate of the atom once it has moved will equal x-2(x+T_x) = x-2x-2T_x = -x-2T_x multiplying by –1 to make x a positive starting point and this gives 2Tx+x Using the same logic for the y and z coordinates gives the coordinates of the atom after the symmetry operation as (2T_x+x, 2T_y+y, 2T_z+z) If T_x = ¼ , T_y = ¼ , Tz = ¼ , then the equation becomes (½+x,½+y,½+z) with the origin being -T_x, -T_y, and -T_z, then the symmetry element will be a point of inversion, where all three coordinates are subject to change providing the origin is located at (-¼,-¼,-¼).

III) –x, -y, -z

This is a point of inversion through (0,0,0)

All three coordinates are inverted by the symmetry operation

IV) –x, -y, ½+z

This is a point of inversion through (0,0,-¼)

½+z follows the logic expressed in II) and the negative signs on the x and y direction show that all three coordinates will be affected by the symmetry, leading to the point of inversion

 

V) x, y, -z

This is a mirror in the z-direction

Only reflection occurs via the z-direction, which will lead to a mirror image since x and y will not be affected

VI) ½+x, -y, z

This is a two fold-rotation parallel to the z-direction with the origin (-¼,0,0)

The origin coordinate for x follow the reasoning in II), Only x and y are affected by the symmetry operation, therefore, rotation will occur in parallel to the unchnged coordinate direction, which is z in this case.

VII) ½+x, -y, -z

This is a point of inversion through (-¼,0,0)

All three coordinates are inverted by the symmetry operation and the origin is determined for the x axis by the same logic as II)

VIII) ½+x, ½+y, z

This is a two-fold rotation parallel to the z-direction with the origin(-¼,-¼,0)

z remains unchanged and x and y will change with respect to the reasoning given in II), leading to a 2-fold rotation.

 

 

 

 

 

 

 

IX) ½-x, ½-y, z

This is a two-fold rotation parallel to the z-direction with the origin (¼,¼,0)

Considering three points T_x,T_y and z as the centre of origin. The distance to the points, x,y,z. By considering the x-coordinate, the difference between x and the starting position T_x will be x-T_x. Once a symmetry operation has taken place, the distance to the new position will similarly be x+T_x. Therefore, the x co-ordinate of the atom once it has moved will equal x-2(x-T_x) = 2T_x-x Using the same logic for the y coordinate gives the coordinates of the atom after the symmetry operation as (2T_x+x, 2T_y+y, z) If T_x= ¼ , Ty = ¼ , then the symmetry operator becomes (½-x,½-y,z) providing the origin is (¼,¼,0)

X) x, ½-y, z

This is a mirror plane in the y axis with the origin (0,-¼,0)

x and z remain unchanged, whilst y will change according to the logic applied in II), resulting in a mirror image in the y-direction via the new origin

General comment: This seems to be your weakest answer. If you have any problems in understanding the model answers, please ask me for more detailed explanations