QuestionMay 7, 2025

The output intensity of an x-ray tube is 3.0mR/mAs when operated at 80 kVp at a distance of100 cm . What would the output intensity be at 200 cm. from the tube? (show your work)

The output intensity of an x-ray tube is 3.0mR/mAs when operated at 80 kVp at a distance of100 cm . What would the output intensity be at 200 cm. from the tube? (show your work)
The output intensity of an x-ray tube is 3.0mR/mAs when operated at 80 kVp at a distance of100 cm . What would the output intensity be at 200 cm. from the tube? (show your work)

Solution
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Answer

0.75 mR/mAs Explanation 1. Identify the relationship The intensity of radiation follows the inverse square law: **I_1/I_2 = (d_2/d_1)^2**. 2. Apply the inverse square law Given I_1 = 3.0 \, \text{mR/mAs} at d_1 = 100 \, \text{cm}, find I_2 at d_2 = 200 \, \text{cm} using I_2 = I_1 \times (d_1/d_2)^2. 3. Calculate the new intensity I_2 = 3.0 \times (100/200)^2 = 3.0 \times (0.5)^2 = 3.0 \times 0.25 = 0.75 \, \text{mR/mAs}.

Explanation

1. Identify the relationship<br /> The intensity of radiation follows the inverse square law: **$I_1/I_2 = (d_2/d_1)^2$**.<br /><br />2. Apply the inverse square law<br /> Given $I_1 = 3.0 \, \text{mR/mAs}$ at $d_1 = 100 \, \text{cm}$, find $I_2$ at $d_2 = 200 \, \text{cm}$ using $I_2 = I_1 \times (d_1/d_2)^2$.<br /><br />3. Calculate the new intensity<br /> $I_2 = 3.0 \times (100/200)^2 = 3.0 \times (0.5)^2 = 3.0 \times 0.25 = 0.75 \, \text{mR/mAs}$.
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