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(B) The intensity of radiation passing through a material of thickness $x$ is given by the formula $I' = I e^{-\mu x}$,where $I$ is the initial intens

Vedclass · Accepted Answer

$12$

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(B) The intensity of radiation passing through a material of thickness $x$ is given by the formula $I' = I e^{-\mu x}$,where $I$ is the initial intensity,$I'$ is the final intensity,and $\mu$ is the absorption coefficient.From the given data,when $x_1 = 36 \, mm$,the intensity becomes $I' = \frac{I}{8}$.Substituting these values: $\frac{I}{8} = I e^{-\mu (36)} \implies e^{36\mu} = 8 = 2^3$.Taking the natural logarithm on both sides: $36\mu = 3 \ln(2) \implies \mu = \frac{3 \ln(2)}{36} = \frac{\ln(2)}{12}$.Now,we need to find the thickness $x_2$ such that the intensity becomes $I' = \frac{I}{2}$.Using the formula: $\frac{I}{2} = I e^{-\mu x_2} \implies e^{\mu x_2} = 2$.Taking the natural logarithm: $\mu x_2 = \ln(2)$.Substituting the value of $\mu$: $(\frac{\ln(2)}{12}) x_2 = \ln(2)$.Solving for $x_2$,we get $x_2 = 12 \, mm$.

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......$mm$

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