The intensity of $γ$ radiation from a given source is I. On passing through 36 mm of lead, it is reduced to I / 8. The thickness of lead, which will reduced the intensity to I/2 will be

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6mm

18mm

12mm

9mm

answer is C.

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Detailed Solution

we know that
$I^{'} = {Ie}^{- μx} or - μX = \ln (I^{'} / I) - μ (36) = \ln (I / 8 I) - μ (x^{'}) = \ln (I / 2 I) \frac{36}{x^{'}} = \frac{3 \ln (2)}{\ln (2)} x^{'} = \frac{36}{3} = 12 mm$

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