Question

A radiographer stands six feet from the x-ray source when performing a portable chest exam and receives an exposure of 2 mGy. If the radiographer performs a repeat exam using the same technical factors standing at a distance of three feet from the source, how much exposure will be received?

Answer 1

Answer:

  I₂ = 8 mG

Explanation:

The intensity of a beam is

          I = P / A

Where P is the emitted power which is 3) 3

           

Let's use index 1 for the initial position of r₁ = 6 ft and 2 for the second position r₂ = 3 ft

          I₁ A₁  = I₂  A₂

           I₂ = I₁ A₁ / A₂

The area of ​​the beam if we assume that it is distributed either in the form of a sphere is

           A₁ = 4π r²

We substitute

            I₂ = I₁ (r₁ / r₂)²

           I₂ = 2 (6/3)²

           I₂ = 2 4

           I₂ = 8 mG

Answer 2

Answer:

The amount of exposure that will be received at 3 ft is 8 mGy

Explanation:

Here, we note that the  amount of radiation exposure of the radiographer is given by the inverse square law. That is the amount of radiation exposure is directly proportional to the inverse square of the distance that is

$\frac{Old \, \, Intensity}{New \, \, \, Intensity} = \frac{(New\, distance)^2}{(Old\, distance)^2} \therefore \frac{2}{New \, \, \, Intensity} = \frac{3^2}{6^2}$

Or New intensity = $2\times \frac{36}{9}$  = 8mGy

Therefore, the amount of exposure that will be received at 3 ft = 8 mGy.