Question

Aluminum is to be used as windows for a cell for X-ray absorption measurements with the Ag $K_{\alpha }$line. The mass absorption coefficient for aluminum at this wavelength is 2.74; its density is $2.70g/c{m}^3$. What maximum thickness of aluminum foil could be used to fabricate the windows if no more than 2.8% of the radiation is to be absorbed by them?

Short answer

Substituting the given the values in the beer-lambert's law gives the value of$x=3.84\times {10}^{-3}cm$

Step-by-step solution

Step 1. Given information

$$\begin{gathered} absorption=2.74 \\ density=2.7g/c{m}^3 \end{gathered}$$

Step 2. Substitute the values in the formula and calculate the value of x

According to beer lambert's law:

$\frac{P_0}{P}=\frac{1}{T}$

WhereTis transmittance

$$\begin{gathered} \frac{P_0}{P}=\frac{1}{0.978} \\ \frac{P_0}{P}=1.02 \end{gathered}$$

Substitute the values in the following formula:

$\ln \frac{P_0}{P}=\rho {\mu }_{M}x$

Where ${\mu }_M$is absorptivity and $\rho$is the density

$$\begin{gathered} \ln (1.02)=2.74\times 2.70\times x \\ x=3.84\times {10}^{-3}cm \end{gathered}$$