Question

LEAD PLATES TOTAL CPM % Radiation Intensity 1 30.94 806 407 2 15 62

how do i find the amount of lead needed to block 99% of the gamma radiation ??

Answer 1

For, I = 806 CPM , radiation intensity = 30.94 %.

then when radiation intensity = 100%, I = 806*100/30.94 = 2605.6 CPM.

** when radiation intensity = 15.62%, I = 15.62*2605.6 / 100 = 406.99 CPM = 407 CPM.

For, 99% blocking of gamma radiation, Io = 100% intensity = 2605.6 CPM.

and I =(100-99)% = 1% intensity = 2605.6 * 0.01 = 26.056 CPM.

From previous calculation, we got penetration factor = = 1.076 cm^-1.

7

so, ln (I/Io) = - *D, where D = thickness of lead plate for 9% radiation blocking.

7

or, ln (26.056 CPM / 2605.6 CPM) = - 1.076 cm^-1 * D cm

or, D = ln (0.01) / (-1.076) = 4.28 cm..

The thickness of plate needed to block 99% of radiation = 4.28 cm.