Question

X-ray diffraction is used to study the structure of crystallized proteins, nucleic acids, and other biological macromolecules. It was actually the technique used to study the structure of DNA in the early 1950s. X-rays of wavelength 0.17 nmnm are to used to study the structure of a protein. One of the maxima in intensity is located at 0.8 radrad from the crystal planes responsible for this maximum.

Answer 1

part 1: c) λ=0.34nm. part 2: a) d=0.118 nm part 3: c) The wavelengths of visible light are too long compared to the atomic spacing.

part 1: Calculating maximum wavelength:

The maximum wavelength of X-ray at the same location is λ2

λ2 = m * λ

where m is the fringe order of interference

here, m = 2 * 0.17

             = 0.34 nm

hence, the X-ray wavelength to give a maximum at the same location is 0.34nm.

part 2: Calculating crystal plane separation:

d is the crystal plane separation.

θ = 0.8 rad

Now, converting radians into degrees.

θ = 0.8 * 180 / Π

  = $45.83^{0}$

Now, calculating d

2 * d * sinθ = m * λ

d = m * λ / 2 * sinθ

  = 1 * 0.17 / 2 * sin($45.83^{0}$)

  = 0.118 nm

part 3:

The wavelength of visible light is 380 to 700 nm which is large to study the structure of proteins.

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Although part of your question is missing, you might be referring to this full question: PART A

Which wavelength of x-ray listed below would give you a maximum at the same location?

a. .26nm

b. .43nm

c. .34nm

d. .60nm

PART B

If this is the first maximum for the X-rays of wavelength 0.17 nm, what is the crystal plane separation of the protein that is responsible?

a. .12nm

b. .24nm

c. .17nm

d. .11nm

PART C

Why can't visible light be used to study the structure of proteins such as this?

a. Visible light reflects off of the surface of proteins.

b. Visible light does not have enough energy.

c. The wavelengths of visible light are too long compared to the atomic spacing.

d. The wavelengths of visible light are too short compared to the atomic spacing.