Answer to Question #121541 in Atomic and Nuclear Physics for Raghuveer

Question

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.

Answer 1

The coefficient of linear attenuation:

$$\mu_\text{l}=\rho\mu_\text{m}.$$

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

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