Question

The intensity of $\gamma$ rays from a source$\left(I_0\right)$ reduces to $\frac{I_0}{8}$ after passing through $48$ mm thick sheet of lead. The thickness of the sheet for obtaining intensity equal to $\frac{I_0}{2}$

• A. $48$ mm
• B. $24$ mm
• C. $16$ mm
• D. $8$ mm

Answer 1

The correct option is C $16$ mm

$I\left(x\right)=I_o{e}^{-\mu x}$

$\frac{I_o}{8}=I_o{e}^{-\mu \times 48}$

$ln8=\mu \times 48$

$\frac{I_o}{2}=I_o{e}^{-\mu x}$

$ln2=\mu \times x=\frac{ln8}{48}\times x$

$\mu =48\times \frac{ln2}{ln8}=48\times \frac{1}{3}=16mm$

So, the answer is option (C).