Question #121541
2. The mass attenuation coefficient of copper is 0.0589 cm2
/g for 1.0-MeV photons. Find required thickness
of the protective copper shield in order to achieve 50 % reduction of X-Ray radiation intensity. The density
of copper is 8.9 g/cm3.

Kindly explain in detail and give the correct solution.
1
Expert's answer
2020-06-12T11:16:46-0400

The coefficient of linear attenuation:


μl=ρμm.\mu_\text{l}=\rho\mu_\text{m}.

The thickness that will provide 50% reduction of X-ray radiation intensity will be


II0=eμlt, t=ln(I0/I)μl=ln(I0/I)ρμm, t=ln(1/0.5)8.90.0589=1.32 cm.\frac{I}{I_0}=e^{-\mu_\text{l}t},\\\space\\ t=\frac{\text{ln}(I_0/I)}{\mu_\text{l}}=\frac{\text{ln}(I_0/I)}{\rho\mu_\text{m}},\\\space\\ t=\frac{\text{ln}(1/0.5)}{8.9\cdot0.0589}=1.32\text{ cm}.

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