Short Answer
A radiographic examination of the small bowel for a 25-cm abdomen was performed doing a 15-min PA; 30-min PA; 60-min PA; and 90-min PA radiograph.What is the total approximate entrance skin exposure if each radiograph was produced at 100 kVp, 25 mA, and 40 in (100 cm)? According to calculations, 100 kVp = 5.6 mR/mA.
Correct Answer:

Verified
Correct Answer:
Verified
Q3: Match the following choices with the correct
Q4: What is the approximate entrance skin exposure
Q5: An effective method of reducing patient exposure
Q6: The approximate entrance skin exposure (ESE) for
Q7: In an effort to minimize patient dose,<br>A)the
Q9: Match the following choices with the correct
Q10: Match the following choices with the correct
Q11: To reduce patient exposure, the _ consistent
Q12: Entrance skin exposure is a/an _ exposure.<br>A)maximum<br>B)minimum<br>C)average<br>D)theoretical
Q13: The conscientious radiographer can reduce the patient