Problem#1
(a) If a chest x ray delivers 0.25 mSv to 5.0 kg of tissue, how many total joules of energy does this tissue receive? (b) Natural radiation and cosmic rays deliver about 0.10 mSv per year at sea level. Assuming an RBE of 1, how many rem and rads is this dose, and how many joules of energy does a 75-kg person receive in a year? (c) How many chest x rays like the one in part (a) would it take to deliver the same total amount of energy to a 75-kg person as she receives from natural radiation in a year at sea level, as described in part (b)?Answer:
equivalent dose (Sv, rem) = RBE × absorbed dose(Gy, rad); 100 rad = 1 Gy
(a) RBE = 1 so 0.25 mSv corresponds to 0.25 mGy.
(b) RBE = 1 so 0.10 mGy = 10 mrad and 10 mrem. (0.10 x 103 J/kg)(75 kg) = 7.5 x 10-3 J
(c) (7.5 x 10-3 J)/(1.2 x 10-3 J) = 6.2. Each chest x ray delivers only about 1/6 of the yearly
background
Problem#2
A person exposed to fast neutrons receives a radiation dose of 200 rem on part of his hand, affecting 25 g of tissue. The RBE of these neutrons is 10. (a) How many rad did he receive? (b) How many joules of energy did this person receive? (c) Suppose the person received the same rad dosage, but from beta rays with an RBE of 1.0 instead of neutrons. How many rem would he have received?
Answer:
The unit for absorbed dose is 1 rad 0.01 J/kg = 0.01 Gy. Equivalent dose in rem is RBE times absorbed dose in rad.
(a) rem = rad x RBE. 200 = x(10) and x = 20 rad.
(b) 1 rad deposits 0.010 J/kg, so 20 rad deposit 0.20 J/kg. This radiation affects 25 g (0.025 kg) of
tissue, so the total energy is (0.025 kg)(0.20 J/kg) = 5.0 x 10-3 J = 5.0 mJ.
(c) RBE = 1 for β-rays, so rem = rad. Therefore 20 rad = 20 rem.
Problem#3
A nuclear chemist receives an accidental radiation dose of 5.0 Gy from slow neutrons (RBE = 4.0). What does she receive in rad, rem, and J/s.
Answer:
The unit for absorbed dose is 1 rad 0.01 J/kg = 0.01 Gy. Equivalent dose in rem is RBE times absorbed dose in rad.
1 rad = 10-2 Gy, so 1 Gy = 100 rad and the dose was 500 rad.
Rem = (rad)(RBE) = (500 rad)(4.0) = 2000 rem. 1Gy = 1 J/kg, so 5.0 J/kg.
Problem#4
To Scan or Not to Scan? It has become popular for some people to have yearly whole-body scans (CT scans, formerly called CAT scans) using x rays, just to see if they detect anything suspicious. A number of medical people have recently questioned the advisability of such scans, due in part to the radiation they impart. Typically, one such scan gives a dose of 12 mSv, applied to the whole body. By contrast, a chest x ray typically administers 0.20 mSv to only 5.0 kg of tissue. How many chest
x rays would deliver the same total amount of energy to the body of a 75-kg person as one whole-body scan?
Answer:
For x rays RBE = 1 so the equivalent dose in Sv is the same as the absorbed dose in J/kg.
One whole-body scan delivers (75 kg)(12 x 10-3 J/kg) = 0.90 J. One chest x ray delivers
(5.0 kg)(0.2 x 10-3 J/kg) = 1.0 x 10-3 J
It takes 0.9 J/(1.0 x 10-3 J) = 900 chest x rays to deliver the same total
Problem#5
Food Irradiation. Food is often irradiated with either x rays or electron beams to help prevent spoilage. A low dose of 5–75 kilorads (krad) helps to reduce and kill inactive parasites, a medium dose of 100–400 krad kills microorganisms and pathogens such as salmonella, and a high dose of 2300–5700 krad sterilizes food so that it can be stored without refrigeration. (a) A dose of 175 krad kills spoilage microorganisms in fish. If x rays are used, what would be the dose in Gy, Sv, and rem, and how much energy would a 220-g portion of fish absorb? (See Table 43.3.) (b) Repeat part (a) if electrons of RBE 1.50 are used instead of x rays.
Answer:
For x rays RBE = 1 and the equivalent dose equals the absorbed dose.
(a) 175 krad = 175 krem = 1.75 kGy = 1.75 kSv. (1.75 x 103 J/kg)(0.220 kg) = 385 J.
(b) 175 krad = 1.75 kGy; (1.50)(175 krad) = 262.5 krem = 2.625 kSv. The energy deposited would be
385 J, the same as in (a).
Problem#6
In an industrial accident a 65-kg person receives a lethal whole-body equivalent dose of 5.4 Sv from x rays. (a) What is the equivalent dose in rem? (b) What is the absorbed dose in rad? (c) What is the total energy absorbed by the person’s body? How does this amount of energy compare to the amount of energy required to raise the temperature of 65 kg of water 0.0100C?
Answer:
1 rem = 0.01 Sv. Equivalent dose in rem equals RBE times the absorbed dose in rad.
1 rad = 0.01 J/kg. To change the temperature of water, Q = mcΔT.
For water, c = 4190 J/kg K.
(a) 5.4 Sv(100 rem/sv) = 540 rem.
(b) The RBE of 1 gives an absorbed dose of 540 rad.
(c) The absorbed dose is 5.4 Gy, so the total energy absorbed is (5.4 Gy)(65 kg) = 351 J. The energy
required to raise the temperature of 65 kg by 0.010 C° is (65 kg)(4190 J/kg K)(0.01 C) = 3 kJ.
Problem#7
A 67-kg person accidentally ingests 0.35 Ci of tritium. (a) Assume that the tritium spreads uniformly throughout the body and that each decay leads on the average to the absorption of 5.0 keV of energy from the electrons emitted in the decay. The half-life of tritium is 12.3 y, and the RBE of the electrons is 1.0. Calculate the absorbed dose in rad and the equivalent dose in rem during one week. (b) The β-1 decay of tritium releases more than 5.0 keV of energy. Why is the average energy absorbed less than the total energy released in the decay?
Answer:
1 rad = 0.01 J/kg. rem = RBE x rad.
λ = ln2/T1/2 = 0.693/(12.3 y x 3.156 x 107 s/y) = 1.785 x 10-9 s.
The number of tritium atoms is
│dN/dt│= λN0
(0.35Ci)(3.70 x 1010 Bq/Ci) = (1.785 x 10-9 s)N0
N0 = 7.254 x 1018 Nuclei
The number of remaining nuclei after one week is
N = N0e-λt = (7.254 x 1018 Nuclei)e-(1.785 x 10-9 s)(7)(24)(3600) = 7.246 x 1018 Nuclei
∆N = N0 – N = 7.8 x 1015 decays.
So the energy absorbed is
∆E = ∆NEγ = (7.8 x 1015)(5000 eV)(1.60 x 10-19 J/eV) = 6.25 J.
The absorbed dose is
6.25 J/67 kg = 0.0932 J/kg = 9.32 rad
Since RBE = 1, then the equivalent dose is 9.32 rem.
Problem#8
In a diagnostic x-ray procedure, 5.00 x 1010 photons are absorbed by tissue with a mass of 0.600 kg. The x-ray wavelength is 0.0200 nm. (a) What is the total energy absorbed by the tissue? (b) What is the equivalent dose in rem?
Answer:
(a) E = hc/λ = (6.63 x 10-34 Js)(2.998 x 108 m/s)/(0.0200 x 10-9 m) = 9.94 x 10-15 J.
The absorbed energy is
(5.00 x 1010 photons)(9.94 x 10-15 J/photon) = 4.97 x 10-4 J = 0.497 mJ
(b) The absorbed dose is
(4.97 x 10-4 J)/(0.600 kg) = 8.28 x 10-4 J/kg = 0.0828 rad.
Since RBE = 1, the equivalent dose is 0.0828 rem.
Post a Comment for "Biological Effects of Radiation Problems and Solutions"