Question

The mass attenuation coefficient for 200 keV x-rays in iron is .1460cm²/g. The density of iron is 7.874 g/cm³. If 10⁴200 keV x-ray photons are incident on an iron plate 1.50 cm

thick, how many x-ray photons will pass through the plate

Answer 1

Mass attenuation coefficient for X-rays in iron is $\frac{\mu }{\rho}=0.1460\,cm^2/g$

Density of iron ${\rho}=7.874\,g/cm^3$

Linear attenuation coefficient for X-rays in iron is $\mu=\frac{\mu}{\rho}{\rho}=0.1460*7.874=1.1496\,cm^{-1}$

Thickness of iron plate $x=1.50\,cm$

Attenuation equation is $I=I_0e^{-\mu x}$

$I=I_0e^{-1.1496*1.5}=0.1783\,I_0$

If ${I_0}=10^4$

$I=0.1783\,I_0=0.1783*10^4=1783$

Number of photons that will pass through the plate are $1783$.