If the radiation dose rate from Cobalt-60 at a distance of 5 centimeters from a small source is 760 mRem per day, at what distance will exposure be reduced by half?
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A: Given: Activity of sample = 9 Ci Half life of sample = 73.8 days
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A: time period (T) = 2 hr rate of deposition = 3.3×10-8 Jsmass of patient (m) = 59 kg radiation type :…
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A: a. The units of absorbed dose are related by 1 rem=10 mSv1 mrem=0.01 mSv Thus X mrem is equivalent…
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Q: 15. An IV infusion is administered to a 120 lb patient by IV infusion at a rate of 1.8 mg/kg/hr. The…
A: Given values: Dose = 1.8 mg/kg/hr * 54.43 kg ≈ 98.0 mg/hr V = 15 L half-life = 8 hours Where: C(t)…
Q: Technetium-99 m has a half life of 4.8 hours. If a patient receives a dose with an activity of 60.2…
A: Given data : Half life (t1/2) = 4.8 hours Dose received (N0) = 60.2 mCi…
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- The linear absorption coefficient for 1-MeV gamma rays in lead is 78 m1. Find the thickness of lead required to reduce by one-fourth the intensity of beam of such gamma rays.If the gamma-ray dose rate is 50 mSv hr-1 at a distance of 1.0 m from a source, what is the dose rate at a distance of 10 m from the source?A Geiger Counter reading of a radio active sample Is initally Cceses 6920 COunts per minute. The same sample gives a reading of 922 COunts per minute 10.9n later. what Is the samples half- 11fe.
- how much dose is absorbed in 30 min in a sphere of water 1cm in diameter, with an infinitesimally small P-32 souce placed at center of the sphere, with 6x105 disintegrations per gram of water? Eb- = 1.71 MeV A) 12 cGy B) 24 cGy C) 29 cGy D) 6cGy E) 59 cGyExperimentally, it is found that natural potassium emits 31 beta particles per second per gram and 3.4 gamma rays per second per gram. With this data, determine the half- life period of K40. (The isotopic concentration of K40 is 0.0118% .)The linear attenuation coefficient for 2.0-MeV gamma rays in water is 4.9 m-1 and 52 m-1 in lead. What thickness of water would give the same shielding for gamma rays as 15 mm of lead?
- So I just need help with the last question where it gives you the 160 The answer for the first question is 1.168 x 10^12 And the answer for the second one is 0.140 , i just need taht last oneWhat is the effective dose received by the prostate gland for an external source in a case where the absorbed dose is 760 J/kg and the radiation is in the form of beta particles. The tissue weighting factor for prostate gland tissue is 0.05.Radiation from a point source follows an inverse-square law in which the amount of radiation received is proportional to 1/d2, where d is distance. If a Geiger counter that is 1 m away from a small source reads 50 counts per minute, what will its reading be a) 2 m from the source b) 3 m from the source.
- A particular radioactive source produces 100 mrad of 2-MeV gamma rays per hour at a distance of 1.0 m. (a) How long could a person stand at this distance before accumulating an intolerable dose of 1 rem? (b) Assuming the gamma radiation is emitted uniformly in all directions, at what distance would a person recieve a dose of 10 mrad/h from this source? dont provode hand written solutionif a normal dose CT scan delivers 700 mrem of radiation, how many CT scans would a patient have to have to feel mild radiation sickness (50 rem)?If a patient receives a dose with an activity of 37.5 mCi of technetium-99m for cardiac imaging, how much radioactivity will be left in the patient’s body 72 hours after injection?