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Problem

A metal with spacing between planes equal to 0.41…
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Problem 102 Medium Difficulty

A diffractometer using $\mathrm{X}$ -rays with a wavelength of 0.2287 nm produced first-order diffraction peak for a crystal angle $\theta=16.21^{\circ} .$ Determine the spacing between the diffracting planes in this crystal.

Answer

0.41 $\mathrm{nm}$

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01:02

Aadit S.

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Video Transcript

in this podcast, we're taking a little bit the lattice structures. More specifically, we're taking a look up Bragg's law so that Bragg's law is used to determine the structure of solid crystals using X ray diffraction. So the equation is N Lander equals to two D Sign theater, where Lambda is the wavelength for X rays. D is the interplay in the distance, which is the distance between two crystal planes. Theta is the angle of diffraction of the defective beam, and N is an integer value. So that's taking a whole numbers with 1234 and so on and so forth. So here we have an equals one. We're going to substitute in some values, so we have one multiplied by 9.2287 nanometers. It's equal to two D sign, 16.21 degrees. So we saw a D, where D is not 0.41 nanometers on. This is the spacing between crystal planes

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Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson

Chemistry

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