69 braggs law gives the condition for diffraction

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69. Bragg’s law gives the condition for diffraction maximum:2dmsinwheredis the spacing of the crystal planes andis the wavelength. The angleismeasured from the surfaces of the planes. For a second-order reflectionm= 2, so9102 0.1210m2.5610m0.26nm.2sin2sin 28md
140770. The angle of incidence on the reflection planes is= 63.8° – 45.0° = 18.8°, and theplane-plane separation isda02 .Thus, using 2dsin=, we getad0220 26021880570sin.sin..nmnm.71. We want the reflections to obey the Bragg condition 2dsin=m, whereis theangle between the incoming rays and the reflecting planes,is the wavelength, andmisan integer. We solve for:FHGIKJFHGIKJsinsin....11990125102 0 252100 2480mdmmmmchch(a) Form= 2 the above equation gives= 29.7°. The crystal should be turned4529.715.3 clockwise.(b) Form= 1 the above equation gives= 14.4°. The crystal should be turned4514.430.6 clockwise.(c) Form= 3 the above equation gives= 48.1°. The crystal should be turned48.1453.1 counterclockwise.(d) Form= 4 the above equation gives= 82.8°. The crystal should be turned82.84537.8 counterclockwise.Note that there are no intensity maxima form> 4, as one can verify by noting thatm/2dis greater than 1 formgreater than 4.72. The wavelengths satisfym= 2dsin= 2(275 pm)(sin 45°) = 389 pm.In the range of wavelengths given, the allowed values ofmarem= 3, 4.(a) The longest wavelength is 389 pm/3 = 130 pm.(b) The associated order number ism= 3.(c) The shortest wavelength is 389 pm/4 = 97.2 pm.(d) The associated order number ism= 4.
CHAPTER 36140873. The sets of planes with the next five smaller interplanar spacings (aftera0) are shownin the diagram that follows.(a) In terms ofa0, the second largest interplanar spacing is0020.7071aa.(b) The third largest interplanar spacing is0050.4472aa.(c) The fourth largest interplanar spacing is00100.3162aa.(d) The fifth largest interplanar spacing is00130.2774aa.(e) The sixth largest interplanar spacing is00170.2425aa.(f) Since a crystal plane passes through lattice points, its slope can be written as the ratioof two integers. Consider a set of planes with slopem/n, as shown in the diagram thatfollows. The first and last planes shown pass through adjacent lattice points along ahorizontal line and there arem– 1 planes between. Ifhis the separation of the first andlast planes, then the interplanar spacing isd = h/m. If the planes make the anglewiththe horizontal, then the normal to the planes (shown dashed) makes the angle= 90° –.The distancehis given byh = a0cosand the interplanar spacing isd = h/m =(a0/m)cos. Since tan=m/n, tan=n/mandcostan.11222mnmThus,dhmamanm0022cos.
140974. (a) We use Eq. 36-14:64R1.2254010mm1.221.3 10rad .5.0mmd(b) The linear separation isD = LR= (150103m) (1.310–4rad) = 20 m.75. Lettingdsin=m, we solve for: dmmmsin( .102500mm / 200)(sin30 )nmwhere1, 2, 3.mIn the visible light rangemcan assume the following values:m1= 4,m2= 5 andm3= 6.(a) The longest wavelength corresponds tom1= 4 with1= 2500 nm/4 = 625 nm.(b) The second longest wavelength corresponds tom2

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Term
Spring
Professor
N/A
Tags
Diffraction, Wavelength, Sin, NM

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