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 0.24 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.

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

a. 0.55 nm
b. 0.33 nm
c. 0.77 nm
d. 0.44 nm

2. If this is the first maximum for the X-rays of wavelength 0.22 nm, what is the crystal plane separation of the protein that is responsible?

a. 0.16 nm
b. 0.31 nm
c. 0.14 nm
d. 0.22 nm

Answer 1

Complete 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 0.22 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.

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

a. 0.55 nm

b. 0.33 nm

c. 0.77 nm

d. 0.44 nm

2. If this is the first maximum for the X-rays of wavelength 0.22 nm, what is the crystal plane separation of the protein that is responsible?

a. 0.16 nm

b. 0.31 nm

c. 0.14 nm

d. 0.22 nm

Answer:

1

  The correct option is D

2

  The correct option is A

Explanation:

From the question we are told that

   The wavelength of X-ray used is  $\lambda = 0.22 \ nm = 0.22 *10^{-9} \ m$

   The  angular displacement is  $\theta = 0.8 \ rad = 0.8 *\frac{180}{3.142} = 45 ^o$

Generally from  Brags equation we have that

      $2d sin \theta = n \lambda$

Here d is the distance of separation between that planes of the protein

and  2d is constant given that the separation does not change

So

      $2d sin (45.84) = n \lambda$    

Given that 2d is constant in order for the angle for $\lambda$ to be the same for $\lambda '$ then  

      $\lambda'$ must be an integral multiple of  $\lambda$

So  

         $\lambda' = 2 * \lambda$

=>      $\lambda' = 2 *0.22$

=>      $\lambda' = 0.44 \ nm$

Considering question 2

From the question we are told that

    The wavelength is  $\lambda = 0.22 \ nm = 0.22 *10^{-9} \ m$

Generally from the above equation we have that

         $d = \frac{\lambda }{2 * sin (\theta )}$

=>      $d = \frac{0.22 * 10^{-9} }{2 * sin (45.84 )}$

=>       $d = 1.6 *10^{-10 } \ m$

=>       $d = 0.16 *10^{-9 } \ m$

=>       $d = 0.16 \ nm$