Question

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

Answer 1

Answer: (B)

$I'=I e^{-\mu x} \Rightarrow x=\frac{1}{\mu} \log _{e} \frac{I}{I'}$
(where $I=$ original intensity, $I'=$ changed intensity)
$36=\frac{1}{\mu} \log _{e} \frac{I}{I / 8}=\frac{3}{\mu} \log _{e} 2$ ...(i)
$x=\frac{1}{\mu} \log _{e} \frac{I}{I / 2}=\frac{1}{\mu} \log _{e} 2$ ...(ii)
From equations (i) and (ii),
$x=12\, mm$