1. A monodirectional beam of 500-kilo electronvolt x-rays having an intensity of 10' x-rays/cm2 - sec is incident on an iron slab 20 cm thick. Calculate: 1. the exposure rate in the absence of the iron 2. the uncollided flux 3. the buildup flux 4. the exposure rate behind the slabStep 1 a) The exposure rate in the absence of the iron is equal to the intentcity Le 107 X-rays/ cm? b) The uncollided flux can be calculated by using Un-collided flux = Incident intensity * (e_-\"d) Explanation: Here, Incident intensity is 1077 x-rays/cm\"2 sec. u is the linear attenuation coefficient for iron, which is 3.3 em -1 d is the thickness of the iron slab, which is 20 cm (a/q) Solving for the value of uncollided flux= 107 x (2.71828_3'33\"1_1\"0) 995743377m"-1 Step 2 c) The build up flux can be solved by using 1ed L Buildup flux = Incident intensity x Explanation: The buildup factor is a measure of the additional attenuation caused by multiple interactions within the material 12 71898 3-3an ' x20cm . _ T 2 Buildup flux = 10" x-rays/ cm* sec x S ten Buildup flux = 32923011.13 Step 3 d) The exposure rate behind the slab can be calculated by Exposure rate behind the slab = Un-collided flux + Buildup flux Explanation: The exposure rate behind the slab is the sum of the un-collided flux and the buildup flux. Exposure rate behind the slab = 32923011.13+995743377 =1028666388