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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0eccdf44-5369-47b5-a163-12ef2fd26cb5%2F1a9251d0-ed12-4a17-a0ea-fa7071813a76%2Fti5moh.jpeg&w=3840&q=75)
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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