Page 408 - Emergency Response Book

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Example:

HVL for Cs-137 is 6.5 mm. of lead. What is the thickness of lead

needed to reduce the dose rate from 40 mR/hr to 2.5 mR/hr ?

n
No. of HVL, 2 =

= 40 / 2.5 = 16

n ln 2 = ln 16  n = ln 16 / ln 2 = 4

Therefore, thickness required = n x No. of HVL

= 4 x 6.5

= 26 mm. of lead

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