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 0.20 nm are to used to study the structure of a protein. One of the maxima in intensity is located at 0.8 rad from the crystal planes responsible for this maximum.
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.

Answer 1

Answer:

PART A: option b. .43nm

PART B: option d. 0.11nm

PART C: option c. The wavelengths of visible light are too long compared to the atomic spacing.

Explanation:

Given data

Wavelength λ = 0.20 nm

Angle θ = 0.8 rad

(a)

wavelength of x-ray to give maximum at the same location

λ₂ = m λ

Here, m = 2 is the interference fringe order.

Substitute the values in the above equation.

λ₂ = 2 × 0.2

    = 0.4 nm

Hence, the wavelength of x-ray to give maximum at the same location is 0.4nm

(b)

The crystal plane separation is equal to d

The value of θ is equal to 0.8 rad.

Convert rad into degree as follows:

0.8 rad = $\frac{180^0}{\pi rad} (\frac{0.8rad}{1})$ = 144°/π = 45.86°

Solve for d, using equation (1) as follows:

2dsinθ = mλ

d = mλ / 2sinθ

d = (1) 0.17 / 2Sin45.86°

d = 0.17 / 1.9065

d = 0.089 nm

(c)

The visible light can not be used to study the structure of proteins because of the high wavelength of the visible light.