If a radiographer receives an exposure of 0.05 millisievert (msv) at a distance of 1.5 feet from the tube of a portable x-ray un
it, what will the exposure be at a distance of 6 feet from the tube?
1 answer:
Answer:
0.003125 msv
Explanation:
= Initial Exposure = 0.05 msv
= Final Exposure
= Initial distance = 1.5 ft
= Final distance = 6 ft
Exposure is inversely related to the distance squared (inverse square law)
![I\propto \dfrac{1}{D}](https://tex.z-dn.net/?f=I%5Cpropto%20%5Cdfrac%7B1%7D%7BD%7D)
So,
![\dfrac{I_1}{I_2}=\dfrac{D_2^2}{D_1^2}\\\Rightarrow I_2=\dfrac{I_1D_1^2}{D_2^2}\\\Rightarrow I_2=\dfrac{0.05\times 1.5^2}{6^2}\\\Rightarrow I_2=0.003125\ msv](https://tex.z-dn.net/?f=%5Cdfrac%7BI_1%7D%7BI_2%7D%3D%5Cdfrac%7BD_2%5E2%7D%7BD_1%5E2%7D%5C%5C%5CRightarrow%20I_2%3D%5Cdfrac%7BI_1D_1%5E2%7D%7BD_2%5E2%7D%5C%5C%5CRightarrow%20I_2%3D%5Cdfrac%7B0.05%5Ctimes%201.5%5E2%7D%7B6%5E2%7D%5C%5C%5CRightarrow%20I_2%3D0.003125%5C%20msv)
The exposure at the given distance is 0.003125 msv
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