Q. The intensity of gamma radiation from a given source is $I$. On passing through $36\ mm$ of lead, it is reduced to $\frac{I}{8}$. The thickness of lead which will reduce the intensity to $\frac{I}{2}$ will be
A) $18\ mm$
B) $12\ mm$
C) $6\ mm$
D) $9\ mm$
Explanation
(b) $I^{'} = I\ e^{- \mu x}\ \Rightarrow x = \frac{1}{\mu}\log_{e}\frac{I}{I^{'}}$ (where $I =$ original intensity, $I^{'} =$ changed intensity)
$36 = \frac{1}{\mu}\log_{e}\frac{I}{I/8} = \frac{3}{\mu}\log_{e}2\ \ \ \ldots(i)$
$x = \frac{1}{\mu}\log_{e}\frac{I}{I/2} = \frac{1}{\mu}\log_{e}2\ \ \ \ \ \ldots(ii)$
From equation (i) and (ii), $x = 12mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.