When an X-ray of wavelength 0.5 , mathring{A} | vedclass

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

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$0.4$

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(B) The intensity of $X$-rays passing through a material is given by the formula: $I = I_0 e^{-\mu x}$,where $I$ is the final intensity,$I_0$ is the initial intensity,$\mu$ is the absorption coefficient,and $x$ is the thickness of the material.Given that the intensity becomes one-fourth of the initial value,we have: $\frac{I}{I_0} = \frac{1}{4}$.Substituting the values into the equation: $\frac{1}{4} = e^{-\mu (3.5 \, mm)}$.Taking the natural logarithm on both sides: $\ln(1/4) = -\mu (3.5 \, mm) \Rightarrow -\ln(4) = -\mu (3.5 \, mm)$.Therefore,$\mu = \frac{\ln(4)}{3.5 \, mm} = \frac{2 \ln(2)}{3.5 \, mm}$.Using $\ln(2) \approx 0.693$: $\mu \approx \frac{2 	imes 0.693}{3.5 \, mm} = \frac{1.386}{3.5 \, mm} \approx 0.396 \, mm^{-1}$.Rounding to the nearest value,we get $\mu \approx 0.4 \, mm^{-1}$.

When an $X$-ray of wavelength $0.5 \, \mathring{A}$ passes through a metal foil of thickness $3.5 \, mm$,its intensity is reduced to one-fourth of its initial value. The absorption coefficient of the metal is ..... $mm^{-1}$.

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