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klemol [59]
3 years ago
12

Bradley gets an x-ray at a radiology clinic that employs its own technologists and radiologists. Would the coder at the clinic r

eport the technical and professional components of this x-ray separately? Would the answer differ if Bradley’s x-ray were sent to an independent radiologist for interpretation? Would any modifiers be used?
Physics
1 answer:
KengaRu [80]3 years ago
8 0

Answer:

Explanation:

If Bradley examination was done and interpreted in the same facility, the radiologist code is used example- procedure code 72100- Radiologic examination, spine, lumbosacral, 2 or 3 views is reported.

if the X-ray was taken by Dr X but Dr X does not read or interpret the image but forward it to the radiologist for initial report, then a 26- modifier is used. E.g A reports by the technologist would be, procedure code 72050-Radiologic examination, spine, cervical, 2 or 3

views or 72050- TC in certain situations and the consulting radiologist would report 72050-26.

if Bradley’s x-ray were sent to an independent radiologist for interpretation, then the procedure code 76140 is used in reporting.

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Calculate the breadth B ( in degrees of 2θ ), due to the small crystal effect alone, of the powder pattern lines of particles of
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Answer:

For differents τ:  

τ = 1000 Å → B = 0.11°

τ = 750 Å → B = 0.15°

τ = 250 Å → B = 0.44°

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Explanation:

To factor B is related to the size of particles, Θ, and λ by the Scherrer equation:  

\tau = \frac{K \lambda}{ B cos(\theta)}

<em>where τ: size of the particles, λ: is the wavelenght of the X-Rays, B: is the line broadening at half the maximum intensity, Θ: angle of incidence and K: is a shape factor with typical value of 0.9 </em>      

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Now, factor B for the diameter of the particles (τ) is:

τ = 1000 Å:          

B = \frac{0.9 \cdot 1.5}{1000 \cdot cos(45)} = 1.91\cdot 10^{-3} rad = 0.109 ^{\circ}                

τ = 750 Å:

B = \frac{0.9 \cdot 1.5}{750 \cdot cos(45)} = 2.54\cdot 10^{-3} rad = 0.146 ^{\circ}

τ = 250 Å:

B = \frac{0.9 \cdot 1.5}{250 \cdot cos(45)} = 7.64\cdot 10^{-3} rad = 0.438 ^{\circ}

For τ = 250 Å, factor B for angles of incidence is:  

Θ = 10°:

B = \frac{0.9 \cdot 1.5}{250 \cdot cos(10)} = 5.48 \cdot 10^{-3} rad = 0.314 ^{\circ}

Θ = 45°:

B = 0.438°  

Θ = 80°:

B = \frac{0.9 \cdot 1.5}{250 \cdot cos(80)} = 0.031 rad = 1.78 ^{\circ}  

Have a nice day!

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