Section IV. RADIATION--PROTECTION
4-14. INTRODUCTION
Radiation protection can be provided in several ways. The factors that determine
the level of protection provided are time, distance, and shielding. Generally speaking,
the time of exposures should be limited to the absolute minimum required for diagnosis
and/or therapy. In other words, ensure that exposure times are kept as short as
possible. Distance from the source of radiation also has an effect on the level of
protection provided. And shielding, which is the intentional use of materials of varying
densities to limit, control, or modify the electromagnetic energy output of an x-ray tube,
is a very useful way of providing protection. To understand the effects of shielding and,
therefore, be able to take advantage of it as a radiation protection tool, it is necessary to
review certain facts about x-ray photon interaction with matter. Photon energy is lost by
many different methods, with photoelectric and Compton effects being the two
predominant in the wavelengths associated with medical x-rays. As x-ray photons
travel through an absorber (the shielding), the amount of reduction, or attenuation, is
determined by three important factors: (1) the energy of the photons, (2) the atomic
mass of the absorbing material, and (3) the thickness of the absorbing material.
4-15. INVERSE SQUARE LAW
Distance from the x-ray source is a highly effective method of reducing the
intensity of an x-ray beam. This can be expressed in terms of the inverse square law,
which states that the x- or gamma-radiation intensity from a point source varies
inversely with the square of the distance from the source. Expressed mathematically,
the inverse square law is:
I1 = (D2)2
I2 (D1)2
where
I1 = intensity at original distance.
I2 = intensity at new distance.
D1 = original distance.
D2 = new distance.
Suppose the intensity of an x-ray beam was 100 R/min at a distance of 2 feet from the
x-ray tube; what would the new intensity be at a distance of 4 feet? By substituting the
data from the above problem into the formula, the new intensity can be found as follows:
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