The attenuation coefficient for aluminum when using a 100 kV…
The attenuation coefficient for aluminum when using a 100 kV x-ray is 0.1925 mm^-1. How much aluminum will it require to reduce a beam’s dose by 50% when using 100 kV x-rays? Given: c = 3.0 ´ 108 m/s (the speed of light) E (in joules) = m (in kg) ´ c2 1 eV = 1.6 ´ 10-19 Joules Planks constant (in j) h = 6.63 ´ 10-34 J-s Planks constant (in eV) h = 4.14 ´ 10-15 eV-s E = h ´ f (in Hz) E(in keV) = 1.24 / l (in nanometers or angstroms) c = l (in meters) ´ f (in Hz) Inverse square law (I = intensity, D = distance) (I(original) / I(new)) = ((D(new))2 / (D(original))2) A= Ao ´ e-[l ´ t] l= decay constant, T½=half life, t=time passed l=ln(2)/ T½ I= Io ´ e-[μ * d] μ=attenuation constant, HVL= Half Value Layer μ=ln(2)/ HVL (1 / T½ (E) )= (1 / T½ (P) ) + (1 / T½ (B)) 1 Bq = 1 disintegration per second (dps) 1 Ci = 3.7 x 1010 Bq 1 mCi = 37 MBq 1 AMU = 1.66 ´ 10-27 kg Mass of e- = 0.00054858 AMU Mass of e- = 9.11 × 10-31 kg Mass of p+ = 1.007276 AMU Mass of p+ = 1.673× 10-27 kg Mass of n0 = 1.008664 AMU Mass of n0 = 1.675× 10-27 kg
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