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(C) The intensity of radiation passing through a material is given by the formula $I = I_0 e^{-\mu t}$,where $I$ is the final intensity,$I_0$ is the i

Vedclass · Accepted Answer

$0.28$

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(C) The intensity of radiation passing through a material is given by the formula $I = I_0 e^{-\mu t}$,where $I$ is the final intensity,$I_0$ is the initial intensity,$\mu$ is the absorption coefficient,and $t$ is the thickness of the material.Given that the intensity is reduced to half $(I = I_0 / 2)$ for a thickness $t = 2.5 \, mm$.Substituting these values into the equation: $I_0 / 2 = I_0 e^{-\mu (2.5)}$.This simplifies to $1/2 = e^{-2.5 \mu}$.Taking the natural logarithm on both sides: $\ln(1/2) = -2.5 \mu$.$-\ln(2) = -2.5 \mu$.$0.693 = 2.5 \mu$.$\mu = 0.693 / 2.5 = 0.2772 \, mm^{-1}$.Rounding to two decimal places,we get $\mu \approx 0.28 \, mm^{-1}$.

The intensity of gamma rays is reduced to half when they pass through a lead sheet of thickness $2.5 \, mm$. The absorption coefficient of lead is ......... $mm^{-1}$.

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