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You have been monitoring your monthly radiation exposure rea…

You have been monitoring your monthly radiation exposure readings and you have noticed they have been high. Normally when you are assisting in the special procedures room during angiography work, you have been standing roughly 2 feet away from the x-ray tube. You bring a radiation detector into the room and discover that the radiation levels at that distance are 15 mR/hr.   How far away from the x-ray tube would you need to stand to have the exposure reduced to < 1 mR/hr?   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

As a radiology manager you are presented the opportunity to…

As a radiology manager you are presented the opportunity to buy replacement lead aprons at a discount price. The new aprons are 2.3 mm (0.23 cm) thick instead of 1.7 mm (0.17 cm) thick for the same price. In general, your technologists, in the fluoroscopy lab, when wearing the old aprons (that are 0.17 cm thick) are getting roughly 30 mR a month under these aprons. What radiation exposure do you anticipate they will be getting after the switch (lead/rubber mixture has a HVL =0.3 cm and the µ = 2.31 cm-1 and the new aprons are 0.23 cm thick and the old were 0.17 cm thick)?   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

The attenuation coefficient (μ) for aluminum when using a 10…

The attenuation coefficient (μ) for aluminum when using a 100 kV x-ray is 0.1925 mm^-1.   What is the half value layer for aluminum 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