1. (5 pts total) A monochromatic x-ray beam is incident on a crystal surface whi
ID: 1697768 • Letter: 1
Question
1. (5 pts total) A monochromatic x-ray beam is incident on a crystal surface which has d = 0.703 nm. The second-order maximum in the reflected beam is found when the angle between the incident beam and the surface is 29.86° .1.a. (2 pts) Using Bragg’s condition, determine the wavelength of the x-rays [in nm].
1.b. (3 pts) Find the scattering angles for the n=1, n=3, and n=4 order diffraction maximums.
2. (5 pts total) Now assume that you are investigating a crystal having unknown lattice spacing, using x-rays of wavelength 0.025 nm and incident on the crystal surface at some angle relative to the surface.
2.a. (1 pt) If you experimentally measured the n=1, n=2, and n=3 scattering angles to be q = 7.181° , 14.48° , and 22.02° respectively, then what is the lattice spacing d?
2.b. (4 pts) What is the highest-order diffraction maximum that can be observed for this wavelength and lattice spacing? (hints: when n becomes too large, you cannot take the inverse-sin of something larger than 1; the answer could be more than 10)
Explanation / Answer
1)
According to Bragg's law
2 d sin = m
Here d = 0.703 nm
= 29.86°
m =2
So the wavelength of the x-rays are
= [2 d sin]/m
= [2(0.703 nm) sin29.86° ]/2
= 0.3500 nm
---------------------------------------------------------
For m=1
= sin-1[m/(2d)]
= sin-1[3(0.3500 nm)/(2(0.703 nm)]
=14.41439o
For m =3
= sin-1[m/(2d)]
= sin-1[3(0.3500 nm)/(2(0.703 nm)]
=48.31389o
For m =4
= sin-1[m/(2d)]
= sin-1[4(0.3500 nm)/(2(0.703 nm)]
=84.70488789o
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