Consider a cubic crystal as shown in the figure below. As this single crystal is

Question

Consider a cubic crystal as shown in the figure below. As this single crystal is rotated in an x-ray spectrometer, many parallel planes of atoms besides AA and BB produce strong diffracted beams. Two such planes are shown in the figure.

Suppose the interplanar spacing is d0 = 33.5 nm.

Determine the interplanar spacing d1.

Determine the interplanar spacing d2.

Suppose the wavelength of the x-rays is l = 2.62 nm. Find the angle (with respect to the surface plane AA) of the n = 1 intensity maxima resulting from scattering from planes with spacing d1.

Find the angle (with respect to the surface plane AA) of the n = 2 intensity maxima resulting from scattering from planes with spacing d1.

Find the angle (with respect to the surface plane AA) of the n = 3 intensity maxima resulting from scattering from planes with spacing d1.

Explanation / Answer

d1 = d0 sin45 = 0.707107*33.5 = 23.688 nm

Bragg's law: 2 d sin(t) = n * lamda

Suppose the wavelength of the x-rays is l = 2.62 nm. Find the angle (with respect to the surface plane AA) of the n = 1 intensity maxima resulting from scattering from planes with spacing d1.

2* 23.688 *sin(t) = 1*2.62

==> t = 3.17 degrees

Find the angle (with respect to the surface plane AA) of the n = 2 intensity maxima resulting from scattering from planes with spacing d1.

2* 23.688 *sin(t) = 2*2.62

==> t = 6.35 degrees

Find the angle (with respect to the surface plane AA) of the n = 3 intensity maxima resulting from scattering from planes with spacing d1.

2* 23.688 *sin(t) = 3*2.62

==> t = 9.55 degrees

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