Question

A radiographic system is using X-ray tube operating at 80 kVp. In order to filter the low energy X-rays, an aluminum (Al) filter of 2.5 mm thickness is used. The Al filter is replaced with a copper (Cu) filter to have the same energy filtered. The mass attenuation coefficients of Al and Cu at 80 kVp are 0.02015 m²/kg and 0.07519 m²/kg, respectively. The densities of Al and Cu are 2699 kg/m³ and 8960 kg/m³, respectively. The thickness of the new Cu filter is \(\underline{\hspace{2cm}}\) mm. (rounded off to two decimal places)

Hint:

For similar energy filtering, use the mass attenuation coefficient and densities to find the equivalent thickness.

Answer 1

Correct Answer: 0.15

The energy attenuation relation is given by: \[ \frac{\mu_{\text{Al}}}{\rho_{\text{Al}}} \times d_{\text{Al}} = \frac{\mu_{\text{Cu}}}{\rho_{\text{Cu}}} \times d_{\text{Cu}} \] Substitute the given values: \[ \frac{0.02015}{2699} \times 2.5 = \frac{0.07519}{8960} \times d_{\text{Cu}} \] Simplify: \[ \frac{0.050375}{2699} = \frac{0.07519}{8960} \times d_{\text{Cu}} \] \[ 0.00001868 = \frac{0.07519}{8960} \times d_{\text{Cu}} \] Solve for \( d_{\text{Cu}} \): \[ d_{\text{Cu}} = \frac{0.00001868 \times 8960}{0.07519} = 0.184\ \text{mm} \] Thus: \[ \boxed{0.18\ \text{mm}} \]