A chest x-ray is taken at 100 kVp, 15 mAs, and a distance of 72 inches, yielding an intensity of 400 mR. If the kVp is reduced to 85 kVp while keeping the distance constant, what will the new intensity be? What mAs is required to maintain the same exposure?

To solve this problem, we'll use two key principles from radiology physics:

1. **The inverse square law**: This states that the intensity of radiation is inversely proportional to the square of the distance from the source. However, since the distance remains constant in this problem, this law will not directly affect the calculation of the new intensity.

2. **The kVp relationship to intensity**: The intensity of the x-ray beam is proportional to the square of the kVp (kilovolt peak). Therefore, if you change the kVp, the intensity will change according to the square of the ratio of the new kVp to the old kVp.

Given:
- Initial intensity (\( I_1 \)) = 400 mR
- Initial kVp (\( \text{kVp}_1 \)) = 100 kVp
- New kVp (\( \text{kVp}_2 \)) = 85 kVp
- Initial mAs (\( \text{mAs}_1 \)) = 15 mAs
- Distance remains constant at 72 inches

### Step 1: Calculate the new intensity (\( I_2 \)) at 85 kVp

The relationship between intensity and kVp is given by:

\[
\frac{I_2}{I_1} = \left(\frac{\text{kVp}_2}{\text{kVp}_1}\right)^2
\]

Substituting the given values:

\[
\frac{I_2}{400} = \left(\frac{85}{100}\right)^2
\]

\[
\frac{I_2}{400} = \left(0.85\right)^2 = 0.7225
\]

\[
I_2 = 400 \times 0.7225 = 289 \text{ mR}
\]

### Step 2: Calculate the new mAs (\( \text{mAs}_2 \)) to maintain the same exposure (intensity)

Since we want to maintain the same exposure (intensity) at 289 mR as the original 400 mR at 15 mAs, we use the direct proportionality of mAs and intensity:

\[
\frac{I_1}{I_2} = \frac{\text{mAs}_2}{\text{mAs}_1}
\]

\[
\frac{400}{289} = \frac{\text{mAs}_2}{15}
\]

\[
\text{mAs}_2 = 15 \times \frac{400}{289} \approx 20.76 \text{ mAs}
\]

### Final Answers:
- The new intensity at 85 kVp is approximately **289 mR**.
- To maintain the same exposure as the initial setting, you would need to increase the mAs to approximately **20.76 mAs**.

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Would you like more details or have any questions? Here are some related questions you might find interesting:

1. How does changing the mA affect image contrast in x-ray imaging?
2. What is the relationship between kVp and the penetration power of x-rays?
3. Why is it important to maintain a specific mAs in radiographic procedures?
4. How does the inverse square law affect x-ray imaging when changing the distance?
5. What are the safety implications of increasing mAs in x-ray imaging?

**Tip:** Always ensure to adjust exposure settings (kVp and mAs) based on patient size and the specific diagnostic requirement to optimize image quality while minimizing radiation exposure.

Understand how to calculate the new x-ray intensity when reducing kVp from 100 to 85 while maintaining a constant distance. Learn how to adjust mAs to maintain exposure with detailed formulas and steps, suitable for radiology and physics students.

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