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(A) The intensity of $X$-rays after passing through a material is given by the formula $I = I_0 e^{-\mu x}$,where $I_0$ is the initial intensity,$\mu$

Vedclass · Accepted Answer

$13.5$

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(A) The intensity of $X$-rays after passing through a material is given by the formula $I = I_0 e^{-\mu x}$,where $I_0$ is the initial intensity,$\mu$ is the attenuation coefficient,and $x$ is the thickness of the material.Given: $\mu = 1.73 \, cm^{-1}$ and $x = 1.156 \, cm$.The ratio of transmitted intensity to initial intensity is $\frac{I}{I_0} = e^{-\mu x}$.Substituting the values: $\frac{I}{I_0} = e^{-(1.73 	imes 1.156)} = e^{-2}$.Since $e \approx 2.718$,$e^2 \approx 7.389$.Thus,$\frac{I}{I_0} = \frac{1}{7.389} \approx 0.1353$.To express this as a percentage,multiply by $100$: $0.1353 	imes 100 = 13.53\%$.Therefore,approximately $13.5\%$ of the $X$-rays will pass through the sheet.

If the attenuation coefficient of aluminum for soft $X$-rays is $1.73 \, cm^{-1}$,what percentage of $X$-rays will pass through an aluminum sheet of thickness $1.156 \, cm$?

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