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4. Width of diffraction maximum. We suppose that in a linear crystal there are identical point scattering centers at every lattice point pm ger. By analogy with (20), the total scattered radiation amplitude will be proportional to F = E exp[-ima Ak]. The sum over M lattice points is = ma, where m is an inte- 1- exp[-iM(a Ak] 1- exp[-i(a · Ak)] F = by the use of the series 1- xM 1- x M-1 %3D m-0 (a) The scattered intensity is proportional to |F|². Show that sin, M(a · Ak) sin (a · Ak) 2. |F| = F*F = (b) We know that a diffraction maximum appears when a ·Ak = 2th, where h is an integer. We change Ak slightly and define e in a · Ak = 27h + e such that e gives the position of the first zero in sin M(a · Ak). Show that e = 27/M, so that the width of the diffraction maximum is proportional to 1/M and can be extremely narrow for macroscopic values of M. The same result holds true for a three-dimensional crystal.

4. Width of diffraction maximum. We suppose that in a linear crystal there are identical point scattering centers at every lattice point pm ger. By analogy with (20), the total scattered radiation amplitude will be proportional to F = E exp[-ima Ak]. The sum over M lattice points is = ma, where m is an inte- 1- exp[-iM(a Ak] 1- exp[-i(a · Ak)] F = by the use of the series 1- xM 1- x M-1 %3D m-0 (a) The scattered intensity is proportional to |F|². Show that sin, M(a · Ak) sin (a · Ak) 2. |F| = F*F = (b) We know that a diffraction maximum appears when a ·Ak = 2th, where h is an integer. We change Ak slightly and define e in a · Ak = 27h + e such that e gives the position of the first zero in sin M(a · Ak). Show that e = 27/M, so that the width of the diffraction maximum is proportional to 1/M and can be extremely narrow for macroscopic values of M. The same result holds true for a three-dimensional crystal.

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4. Width of diffraction maximum. We suppose that in a linear crystal there are identical point scattering centers at every lattice point pm ger. By analogy with (20), the total scattered radiation amplitude will be proportional to F = E exp[-ima Ak]. The sum over M lattice points is = ma, where m is an inte- 1- exp[-iM(a Ak] 1- exp[-i(a · Ak)] F = by the use of the series 1- xM 1- x M-1 %3D m-0 (a) The scattered intensity is proportional to |F|². Show that sin, M(a · Ak) sin (a · Ak) 2. |F| = F*F = (b) We know that a diffraction maximum appears when a ·Ak = 2th, where h is an integer. We change Ak slightly and define e in a · Ak = 27h + e such that e gives the position of the first zero in sin M(a · Ak). Show that e = 27/M, so that the width of the diffraction maximum is proportional to 1/M and can be extremely narrow for macroscopic values of M. The same result holds true for a three-dimensional crystal.
Transcribed Image Text:4. Width of diffraction maximum. We suppose that in a linear crystal there are identical point scattering centers at every lattice point pm ger. By analogy with (20), the total scattered radiation amplitude will be proportional to F = E exp[-ima Ak]. The sum over M lattice points is = ma, where m is an inte- 1- exp[-iM(a Ak] 1- exp[-i(a · Ak)] F = by the use of the series 1- xM 1- x M-1 %3D m-0 (a) The scattered intensity is proportional to |F|². Show that sin, M(a · Ak) sin (a · Ak) 2. |F| = F*F = (b) We know that a diffraction maximum appears when a ·Ak = 2th, where h is an integer. We change Ak slightly and define e in a · Ak = 27h + e such that e gives the position of the first zero in sin M(a · Ak). Show that e = 27/M, so that the width of the diffraction maximum is proportional to 1/M and can be extremely narrow for macroscopic values of M. The same result holds true for a three-dimensional crystal.
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