1 if the radiation intensity of a point source at 05m...

1) If the radiation intensity of a point source at 0.5m measures 36 mR/hr, at what distance will the intensity be halved? 2) If the HVL for I131 in lead is 0.3 cm, what is the minimum thickness of lead that is required to reduce the exposure rate of an I131 source from 12 mR/hr to less than 2 mR/hr? 3) If the HVL of Co60 in lead is 1.25 cm, what percentage of Co60 photons is absorbed from a Co60 source shielded with 2.5 cm of lead? 4) If a point source produces an exposure rate of 30 mR/hr at a distance of 15 cm, what is the exposure rate at 40 cm from the source?

Transcript

00:01 Hi, here for the first question we need to half the value of radiation intensity.

00:06 So here for the first one we know that relation between i1 and i2 can be written as distance 2 upon distance 1 whole square.

00:18 So here in our case now further we are given that i1 is equal to 18, i2 is equal to 36, d2 we need to calculate d1 is equal to 0 .5.

00:32 So substituting the value here we have 18 upon 36 is equal to distance d2 divided by 0 .5 whole square.

00:44 So here now solving this we can say that here in our case the value of distance d2 equals to 0 .5.

00:56 So here we can say that we got our first solution as this is the value of the distance.

01:02 Now for the second part of the question here we need to find the value of half layers and the lid shieldings.

01:10 So here in our case now we are given that the value of half layer value hvl equals to 0 .3 centimeter and further we are given that the value of initial exposure rate initial exposure rate equals to 12 and desired exposure rate is less than 2.

01:50 So here in our case now we can use the hvl formula to find the value of thickness of the lid.

01:58 So here in our case now we have value of initial intensity divided by final intensity equals to 2.

02:10 So here in our case now further we can say that lid thickness divided by hvl equals to log 2 base initial intensity divided by final intensity.

02:25 So here in our case now substituting the values further we can say that here we have thickness t divided by 0 .3 equals to log 2 base 12 upon 2.

02:37 So solving this for thickness we can say that here in our case the value of thickness t is equal to 0 .7755 centimeter.

02:50 So this is our second solution...

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