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The intensity of gamma radiation from a given source is $I$. On passing through $36 \mathrm{~mm}$ of lead, it is reduced to $I / 8$. The thickness of lead which will reduce the intensity to $I / 2$ will be
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Verified Answer
The correct answer is:
$12 \mathrm{~mm}$
$$
\begin{aligned}
& \text { } \because I=I_0 e^{-k x} \Rightarrow \frac{I}{I_0}=e^{-k x} \\
& \therefore \quad \ln \left(\frac{I}{I_0}\right)=-k x
\end{aligned}
$$
In first case
$$
\begin{aligned}
& \ln \left(\frac{1}{8}\right)=-k \times 36 \\
& \ln \left(2^{-3}\right)=-k \times 36 \\
& \text { or } 3 \ln 2=k \times 36
\end{aligned}
$$
In second case, $\ln \left(\frac{1}{2}\right)=-k \times x$
or $\ln \left(2^{-1}\right)=-k x$
or $\ln 2=k x$
From (i) and (ii)
$$
3 \times(k x)=k \times 36
$$
$$
\text { or } x=12 \mathrm{~mm} \text {. }
$$
\begin{aligned}
& \text { } \because I=I_0 e^{-k x} \Rightarrow \frac{I}{I_0}=e^{-k x} \\
& \therefore \quad \ln \left(\frac{I}{I_0}\right)=-k x
\end{aligned}
$$
In first case
$$
\begin{aligned}
& \ln \left(\frac{1}{8}\right)=-k \times 36 \\
& \ln \left(2^{-3}\right)=-k \times 36 \\
& \text { or } 3 \ln 2=k \times 36
\end{aligned}
$$
In second case, $\ln \left(\frac{1}{2}\right)=-k \times x$
or $\ln \left(2^{-1}\right)=-k x$
or $\ln 2=k x$
From (i) and (ii)
$$
3 \times(k x)=k \times 36
$$
$$
\text { or } x=12 \mathrm{~mm} \text {. }
$$
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