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Unformatted text preview: "'Jl—IJI.IIIJ I' u 3.30. (3) Sketch. in a cubic unit cell, [100] and [210] di
rections. (1:) Use a trigonometric calculation to de
termine the angle between these directions. (1:) Use
Equation 3.3 to determine this angle 5“: Eliam
_ _ hf;
,, r “4... a
:4". 4... (V1): 36.5. (a) 5: art as 2+o+a WW7: teatc (as 1?_5_— '' 26.5. 3.35. A useful rule of thumb for the cubic system it than given
[an] Citation in the normal In the (m) 9% Using
this rule and Equation 3,}. ImamLine which members of
the (110} family of dixeeﬁens lie mum: the (111} plane.
(Hints The dot product of two perpendicular vectors is
zero.) 0:01. 4.4. ﬁlial honu! f. (w) __
M [M] m‘flJMn 3m («f/”Ag" mag . _ [TaJ, [I Ta 1, £ rm]. Us 7]. [M], a.) £977] _ 3.44. The four—digit notation system (MillerBrant in—
dices) inncanted for planes in the hexagonal sync!!!
can also be used for deseribingcrystal directions. 'In
a hexagqnal uni_t eeil sketch (8)1118 [0001] direction.
and (b)'1h:[1120] direction. 3.78. The diffraction peaks Iabeled in Figure 3:39 corre
spond to the reﬂection rules for'an foe meta] (h. k. I
unmixed. as shown in Table 3.4). What would be the
(Mr!) indices for the three lowest diffraction anye
peaks for a bee metal? me, we run" M'ﬂé {4: 5.4%.. :
' kWh[Inb nunJ“ Secn4, M a... Chu‘Jhc, E’ml‘m 3.5% 3.6:
.3151 9 fig1f an} ‘5‘]! :97 W?“
(Lu): (Ito) (200) £425
1.44;): a 2 4‘
A‘JPd': A 4' 6 3.79 Using the result of Problem 3.73. calculate the diﬁractinn
angles (29) for the ﬁrst three peaks in the diﬁmctiun pu
tem of calFe powder using Canradiation (1 = 0.1542 . nm)
For catFe, ﬂaw4r}: or a: (4”?)52
= (4WJKMJJh. ;
= 0.236 a... d = 0.336 to": 0”‘un
{n+0 2 J . 0.23611  Gillian1 .
an gﬁ a .0. 143:... d 0.216.... _ c.2361... gull/1%)“  1’6 Indra... 0.154230”
armada...  0.29: “In 9m=mm  2.3.4 « res)”. 44.2 eme'Hvfij—jgi : 32.5' r (30)“, 65.33
I n a“ .  0.154:  .. '_ o
93.. ““253: 4“" " 3"»? ”5 ...
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 Fall '09
 Stroeve

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