Question

The thickness of lead, which will reduce the intensity to 1/2 will be :

• A. 6 mm
• B. 9 mm
• C. 18 mm
• D. 12 mm

Answer 1

The correct option is D 12 mm

$\begin{array}{c}\text{On passing through lead, the intensity of radiatfon.}\\ \text{decreases with, thirckness by formula}\\ I=I_0{e}^{-\mu x}\end{array}$

$\begin{array}{c}\text{first case}\\ \begin{array}{c}x=36mm,I=I_0/8\\ \frac{I_0}{8}=I_0{e}^{-\mu 36}\\ {\log }_{e}8=36\mu \Rightarrow 3{\log }_{e}2=36\mu \end{array}\end{array}$

$\begin{array}{c}\text{Second case}\\ I=I_0/2\\ \frac{I_0}{2}=\pi e^{-\mu x}\\ {\log }_{e}2=\mu x\\ \frac{36\mu }{3}=\mu x\\ x=12mm\end{array}$