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textbook:nrctextbook:chapter18 [2025-02-13 10:25]
Merja Herzig 		textbook:nrctextbook:chapter18 [2025-05-05 13:25] (current)
Merja Herzig
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-1. a) 0.0118% of natural potassium is radioactive <sup>40</sup>K isotope the half-life of which is 1.27×10<sup>9</sup> years. What is the <sup>40</sup>K activity in a human body in which the amount of potassium is 125 g? Molar mass of potassium is 39.1 g/mol.+1. a) 0.0118% of natural potassium is radioactive <sup>40</sup>K [[textbook:nrctextbook:chapter2#isotope|isotope]] the [[textbook:nrctextbook:chapter6#half_life|half-life]] of which is 1.27×10<sup>9</sup> years. What is the <sup>40</sup>K [[textbook:nrctextbook:chapter6#activity|activity]] in a human body in which the amount of potassium is 125 g? Molar mass of potassium is 39.1 g/mol.
  
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-b) What is the uncertainty of the <sup>40</sup>K activity calculated above when the uncertainty of the potassium mass (125 g) is 15%? Use the uncertainty propagation law.+b) What is the [[textbook:nrctextbook:chapter14#standard_deviation_of_activity|uncertainty]] of the <sup>40</sup>K [[textbook:nrctextbook:chapter6#activity|activity]] calculated above when the uncertainty of the potassium mass (125 g) is 15%? Use the uncertainty propagation law.
  
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-2. a) The activity of a radionuclide decreased to one per mil of the initial value. What is the half-life of the radionuclide?+2. a) The [[textbook:nrctextbook:chapter6#activity|activity]] of a radionuclide decreased to one per mil of the initial value in seven days. What is the [[textbook:nrctextbook:chapter6#half_life|half-life]] of the radionuclide?
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-b) What is the activity of 8 MBq of <sup>90</sup>Sr after 42 years? Half-life is 28.8 years.+b) What is the [[textbook:nrctextbook:chapter6#activity|activity]] of 8 MBq of <sup>90</sup>Sr after 42 years? [[textbook:nrctextbook:chapter6#half_life|Half-life]] is 28.8 years.
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-3. What is the <sup>14</sup>C/C ratio in a carbon sample the specific activity of which is 50.1 dpm/g? Half-life of <sup>14</sup>C is 5730 years.+3. What is the <sup>14</sup>C/C ratio in a carbon sample the [[textbook:nrctextbook:chapter6#specific_activity|specific activity]] of which is 50.1 dpm/g? [[textbook:nrctextbook:chapter6#half_life|Half-life]] of <sup>14</sup>C is 5730 years.
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-4. How many impulses should be recorded from a radioactive sample to observe a standard deviation of the number of impulses lower than 0.5%?+4. How many impulses should be recorded from a radioactive sample to observe a of the number of impulses lower than 0.5%?
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-5. The counting efficiency of a whole-body counting system is 1.35% at 935 keV gamma energy. The background count rate at this energy was observed to be 5798 pulses in 60 minutes. A person contaminated with <sup>115</sup>Cd was measured for 15 minutes and the number of pulses at 935 keV was 3987. What was the <sup>115</sup>Cd activity in the body of this person? <sup>115</sup>Cd emits gamma rays of 935 keV energy with 1.90% intensity.+5. The [[textbook:nrctextbook:chapter8#counting_efficiency|counting efficiency]] of a whole-body counting system is 1.35% at 935 keV gamma energy. The [[textbook:nrctextbook:chapter14#background|background count rate]] at this energy was observed to be 5798 pulses in 60 minutes. A person contaminated with <sup>115</sup>Cd was measured for 15 minutes and the number of pulses at 935 keV was 3987. What was the <sup>115</sup>Cd [[textbook:nrctextbook:chapter6#activity|activity]] in the body of this person? <sup>115</sup>Cd emits gamma rays of 935 keV energy with 1.90% intensity.
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-6. Calculate the uncertainty of the human body activity determined in exercise 5. Use uncertainty propagation law. You do not need to take into account uncertainties of intensity, counting efficiency and counting time. Express the result as the activity with its uncertainty.+6. Calculate the [[textbook:nrctextbook:chapter14#standard_deviation_of_activity|uncertainty]] of the human body activity determined in exercise 5. Use uncertainty propagation law. You do not need to take into account uncertainties of intensity, counting efficiency and counting time. Express the result as the activity with its uncertainty.
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-7. Marie and Pierre Curie used 2 tons of pitchblende (75% U<sub>3</sub>O<sub>8</sub>) as a starting material to isolate radium. What was the amount of <sup>226</sup>Ra (mg) in radioactive equilibrium with the amount of uranium initially present. For simplicity we assume that all uranium was only <sup>238</sup>U the molar mass of which is 238.03 g/mol and its half-life is 4.5×10<sup>9</sup> years. The half-life of <sup>226</sup>Ra is 1600 years.+7. Marie and Pierre Curie used 2 tons of pitchblende (75% U<sub>3</sub>O<sub>8</sub>) as a starting material to isolate radium. What was the amount of <sup>226</sup>Ra (mg) in [[textbook:nrctextbook:chapter6#activity_equilibrium|radioactive equilibrium]] with the amount of uranium initially present. For simplicity we assume that all uranium was only <sup>238</sup>U the molar mass of which is 238.03 g/mol and its [[textbook:nrctextbook:chapter6#half_life|half-life]] is 4.5×10<sup>9</sup> years. The half-life of <sup>226</sup>Ra is 1600 years.
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 1. A sample with an activity of 55.0 kBq was measured with a GM counter and number of  1. A sample with an activity of 55.0 kBq was measured with a GM counter and number of 
-observed net counts was 280 000. Another sample with an activity of 55.0 Bq, obtained by diluting the first sample by a factor of 1000, was then measured with the same GM counter in identical conditions. In this case the number of pulses was 350. Calculate the counting efficiency of the counting system and dead-time of the GM detector.+observed [[textbook:nrctextbook:chapter14#net_count_rate|net counts]] was 280 000. Another sample with an activity of 55.0 Bq, obtained by diluting the first sample by a factor of 1000, was then measured with the same GM counter in identical conditions. In this case the number of pulses was 350. Calculate the [[textbook:nrctextbook:chapter8#counting_efficiency|counting efficiency]] of the counting system and [[textbook:nrctextbook:chapter8# dead_time |dead-time]] of the GM detector.
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-2. Calculate the uncertainties of counting efficiency and dead-time observed in exercise 1, using uncertainty propagation law, for a case where the uncertainty of both samples was 2%. Express the results as the counting efficiency and the dead-time with their uncertainties.+2. Calculate the uncertainties of [[textbook:nrctextbook:chapter8#counting_efficiency|counting efficiency]] and [[textbook:nrctextbook:chapter8# dead_time |dead-time]] observed in exercise 1, using uncertainty propagation law, for a case where the uncertainty of both samples was 2%. Express the results as the counting efficiency and the dead-time with their uncertainties.
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-3. An unknown radionuclide was measured with a GM counter having a dead-time of 32 µs. The observed net count rate was 280 000 imp/min on March 15<sup>th</sup> and 123 imp/min on November 10<sup>th</sup>  of the same year. We assume the counting efficiency having been constant. Calculate the half-life of the radionuclide.+3. An unknown radionuclide was measured with a GM counter having a [[textbook:nrctextbook:chapter8# dead_time |dead-time]] of 32 µs. The observed [[textbook:nrctextbook:chapter14#net_count_rate|net count rate]] was 280 000 imp/min on March 15<sup>th</sup> and 123 imp/min on November 10<sup>th</sup> of the same year. We assume the [[textbook:nrctextbook:chapter8#counting_efficiency|counting efficiency]] having been constant. Calculate the [[textbook:nrctextbook:chapter6#half_life|half-life]] of the radionuclide.
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 ==== 18.3.1. Exercises ==== ==== 18.3.1. Exercises ====
  
-1. Molybdenum was irradiated with thermal neutrons to produce radioactive <sup>99</sup>Mo (t<sub>1/2</sub> = 2.75 d)isotope from stable <sup>98</sup>Mo isotope. <sup>99</sup>Mo decays by beta decay to <sup>99</sup>Tc (t<sub>1/2</sub> = 2.13 x 10<sup>5</sup> a) which further decays by beta decay to stable <sup>99</sup>Ru. After irradiation molybdenum was separated from the target material; the <sup>99</sup>Mo activity in the separated molybdenum was 100 kBq. What were the <sup>99</sup>Mo and <sup>99</sup>Tc activities after **a) 1 h, b) 3d, c) 1a** ?+1. Molybdenum was irradiated with [[textbook:nrctextbook:chapter15#thermal_neutron|thermal neutrons]] to produce radioactive <sup>99</sup>Mo (t<sub>1/2</sub> = 2.75 d) [[textbook:nrctextbook:chapter2#isotope|isotope]] from stable <sup>98</sup>Mo isotope. <sup>99</sup>Mo decays by beta decay to <sup>99</sup>Tc (t<sub>1/2</sub> = 2.13 x 10<sup>5</sup> a) which further decays by [[textbook:nrctextbook:chapter5#5.3._beta_decay_processes|beta decay]] to stable <sup>99</sup>Ru. After irradiation molybdenum was separated from the [[textbook:nrctextbook:chapter15#target_nucleus|target]] material; the <sup>99</sup>Mo activity in the separated molybdenum was 100 kBq. What were the <sup>99</sup>Mo and <sup>99</sup>Tc activities after **a) 1 h, b) 3d, c) 1a** ?
 Use equation: Use equation:
 $$A_2 = \frac{\lambda_2 \lambda_1}{\lambda_2 - \lambda_1} N_1^0 \left( 2^{-\frac{t}{t_1}} - 2^{-\frac{t}{t_2}} \right)$$ $$A_2 = \frac{\lambda_2 \lambda_1}{\lambda_2 - \lambda_1} N_1^0 \left( 2^{-\frac{t}{t_1}} - 2^{-\frac{t}{t_2}} \right)$$
  
-2. A fish sample of 2.0 g weight was irradiated in a reactor for 2.0 hours with a neutron flux of 5.0×10<sup>17</sup> (φ). Twenty days after the irradiation the activity of generated <sup>203</sup>Hg activity wasmeasured to be 226 Bq. Calculate the mercury concentration of the fish as mg/kg. The cross section (σ) of the reaction <sup>202</sup>Hg(n,γ)<sup>203</sup>Hg is 380 fm<sup>2</sup>, the half-life of <sup>203</sup>Hg is 47 days, the isotopic abundance of <sup>202</sup>Hg is 29.8% and the atomic weight of mercury is 200.6 .+2. A fish sample of 2.0 g weight was irradiated in a [[textbook:nrctextbook:chapter16#radionuclide_production_reactors|reactor]] for 2.0 hours with a neutron flux of 5.0×10<sup>17</sup> (φ). Twenty days after the irradiation the activity of generated <sup>203</sup>Hg was measured to be 226 Bq. Calculate the mercury concentration of the fish as mg/kg. The [[textbook:nrctextbook:chapter15#cross_section|cross section]] (σ) of the reaction <sup>202</sup>Hg(n,γ)<sup>203</sup>Hg is 380 fm<sup>2</sup>, the [[textbook:nrctextbook:chapter6#half_life|half-life]] of <sup>203</sup>Hg is 47 days, the [[textbook:nrctextbook:chapter2#isotope|isotopic abundance]] of <sup>202</sup>Hg is 29.8% and the [[textbook:nrctextbook:chapter2#atomic_weight|atomic weight]] of mercury is 200.6 .
 Use equation: Use equation:
  
 $$A_\tau = \sigma \phi N_1 \left( 1 - 1^{-\frac{t_S}{T}} \right) 2^{-\frac{t}{T}}$$ $$A_\tau = \sigma \phi N_1 \left( 1 - 1^{-\frac{t_S}{T}} \right) 2^{-\frac{t}{T}}$$
  
-3. To determine the neutron flux of a reactor 10 mg of copper was irradiated for 4.0 hours. 25 hours and 36 minutes after the irradiation was terminated the <sup>64</sup>Cu activity was 70 MBq. The cross section of the reaction <sup>63</sup>Cu(n,γ)<sup>64</sup>Cu is 4.51 barn , the isotopic abundance of <sup>63</sup>Cu is 69.1%, the atomic weight of copper is 63.5 g/mol and the half-life of <sup>64</sup>Cu is 12.8 hours. What is the neutron flux of the reactor.+3. To determine the neutron flux of a [[textbook:nrctextbook:chapter16#radionuclide_production_reactors|reactor]],  10 mg of copper was irradiated for 4.0 hours. 25 hours and 36 minutes after the irradiation was terminated the <sup>64</sup>Cu activity was 70 MBq. The [[textbook:nrctextbook:chapter15#cross_section|cross section]] of the reaction <sup>63</sup>Cu(n,γ)<sup>64</sup>Cu is 4.51 [[textbook:nrctextbook:chapter15#barn|barn]], the [[textbook:nrctextbook:chapter2#isotope|isotopic abundance]] of <sup>63</sup>Cu is 69.1%, the [[textbook:nrctextbook:chapter2#atomic_weight|atomic weight]] of copper is 63.5 g/mol and the [[textbook:nrctextbook:chapter6#half_life|half-life]] of <sup>64</sup>Cu is 12.8 hours. What is the neutron flux of the reactor?
  
-4. <sup>131</sup>Ba (t<sub>1/2</sub> = 11.5 d) decays by positron emission to <sup>131</sup>Cs (t1/2 = 9.7 d) and the latter further to stable <sup>131</sup>Xe. Barium is separated from a sample containing both nuclides. What are the <sup>131</sup>Ba and <sup>131</sup>Cs activities in the separated barium fraction after 1, 10 and 100 days when the initial <sup>131</sup>Ba activity was 100 kBq. Plot, with excel or other suitable program, the activities of both nuclides for the time +4. <sup>131</sup>Ba (t<sub>1/2</sub> = 11.5 d) decays by [[textbook:nrctextbook:chapter5#positron_decay|positron emission]] to <sup>131</sup>Cs (t1/2 = 9.7 d) and the latter further to stable <sup>131</sup>Xe. Barium is separated from a sample containing both [[textbook:nrctextbook:chapter2#nuclide|nuclides]]. What are the <sup>131</sup>Ba and <sup>131</sup>Cs [[textbook:nrctextbook:chapter6#activity|activities]] in the separated barium fraction after 1, 10 and 100 days when the initial <sup>131</sup>Ba activity was 100 kBq. Plot, with excel or other suitable program, the activities of both nuclides for the time 
 period up to 100 days. When the nuclide activities are identical? When the <sup>131</sup>Cs activity has its maximum? period up to 100 days. When the nuclide activities are identical? When the <sup>131</sup>Cs activity has its maximum?
  
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 ==== 18.4.1. Exercises ==== ==== 18.4.1. Exercises ====
  
-1. A medical doctor determines the blood volume of a patient by injecting 0.15 ml of <sup>99m</sup>TcO<sub>4</sub> (t<sub>½</sub>= 6.01 hours) solution into the patient’s blood vessel. The activity concentration of the solution was 30 MBq/mL at the time of injection. After 4 hours a sample of the blood is taken and its activity concentrations is measured to be 610 Bq/mL. Calculate the total blood volume of the  +1. A medical doctor determines the blood volume of a patient by injecting 0.15 ml of <sup>99m</sup>TcO<sub>4</sub> (t<sub>½</sub>= 6.01 hours) solution into the patient’s blood vessel. The [[textbook:nrctextbook:chapter6#activity_concentration|activity concentration]] of the solution was 30 MBq/mL at the time of injection. After 4 hours a sample of the blood is taken and its activity concentrations is measured to be 610 Bq/mL. Calculate the total blood volume of the patient.
-patient.+
  
-2. Calculate the uncertainty of the blood volume determined in exercise 1. The uncertainty of the injection volume was 0.01 ml and the relative uncertainties of the initial and final <sup>99m</sup>Tc activities were 5% (uncertainty of the counting time is very small and is not taken into account). Calculate the uncertainty by using uncertainty propagation law and express the result as the total blood volume with its uncertainty.+2. Calculate the [[textbook:nrctextbook:chapter14#standard_deviation_of_activity|uncertainty]] of the blood volume determined in exercise 1. The uncertainty of the injection volume was 0.01 ml and the relative uncertainties of the initial and final <sup>99m</sup>Tc activities were 5% (uncertainty of the counting time is very small and is not taken into account). Calculate the uncertainty by using uncertainty propagation law and express the result as the total blood volume with its uncertainty.
  
-3. <sup>226</sup>Radium (M = 226.025 g/mol) decays to <sup>222</sup>Rn with a half-life of 1600 years. What is the volume of the <sup>222</sup>Rn gas generated from 42 kg of Ra (T=25 °C ja p = 1 atm)?+3. <sup>226</sup>Radium (M = 226.025 g/mol) decays to <sup>222</sup>Rn with a [[textbook:nrctextbook:chapter6#half_life|half-life]] of 1600 years. What is the volume of the <sup>222</sup>Rn gas generated from 42 kg of Ra (T=25 °C ja p = 1 atm)?
  
 ==== 18.4.2. Solutions ==== ==== 18.4.2. Solutions ====
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 $$= \sqrt{ $$= \sqrt{
-\left(\frac{0.15 \, \text{mL} \cdot 2^{-\frac{4h}{6.01h}} \cdot 0.05 \cdot 30 \cdot 10^6 \, \text{Bq/mL}}{610 \, \text{Bq/mL}}\right)^2 +  +\begin{array}{l} 
-\left(-\frac{30 \cdot 10^6 \, \text{Bq/mL} \cdot 0.15 \, \text{mL} \cdot 2^{-\frac{4h}{6.01h}} \cdot 0.05 \cdot 610 \, \text{Bq/mL}}{(610 \, \text{Bq/mL})^2}\right)^2 +\left(\frac{0.15 \, \text{mL} \cdot 2^{-\frac{4h}{6.01h}} \cdot 0.05 \cdot 30 \cdot 10^6 \, \text{Bq/mL}}{610 \, \text{Bq/mL}}\right)^2 + \\ 
-} \\ +\left(-\frac{30 \cdot 10^6 \, \text{Bq/mL} \cdot 0.15 \, \text{mL} \cdot 2^{-\frac{4h}{6.01h}} \cdot 0.05 \cdot 610 \, \text{Bq/mL}}{(610 \, \text{Bq/mL})^2}\right)^2 + \\
-+ \sqrt{+
 \left(\frac{30 \cdot 10^6 \, \text{Bq/mL} \cdot 2^{-\frac{4h}{6.01h}} \cdot 0.01 \, \text{mL}}{610 \, \text{Bq/mL}}\right)^2 \left(\frac{30 \cdot 10^6 \, \text{Bq/mL} \cdot 2^{-\frac{4h}{6.01h}} \cdot 0.01 \, \text{mL}}{610 \, \text{Bq/mL}}\right)^2
 +\end{array}
 }$$ }$$
  
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 $$\rightarrow V_{\text{blood}} = (4.7 \pm 0.5) \, \text{L}$$ $$\rightarrow V_{\text{blood}} = (4.7 \pm 0.5) \, \text{L}$$
 +
 +
 +**3.**
 +
 +**first the number of generated <sup>222</sup>Rn atoms is calculated:**
 +
 +$$N_{\text{Rn}} = N_{Q\text{Ra}} - N_{\text{Ra}} = N_{Q\text{Ra}} - N_{Q\text{Ra}} \cdot 2^{-\frac{t}{t_{1/2}}} = \frac{m_{\text{Ra}}}{M_{\text{Ra}}} \cdot N_A \cdot \left(1 - 2^{-\frac{t}{t_{1/2}}}\right)$$
 +
 +**using the ideal gas law the volume can be calculated:**
 +
 +$$pV = nRT \rightarrow V = \frac{nRT}{N_A p} = \frac{4.846678 \cdot 10^{22} \cdot 0.08206 \, \text{(L} \cdot \text{atm)/(mol} \cdot \text{K)} \cdot 298.15 \, \text{K}}{6.022 \cdot 10^{23} \, \text{1/mol} \cdot 1 \, \text{atm}} \approx 1.97 \, \text{L}$$
 +
 +
  
textbook/nrctextbook/chapter18.1739438732.txt.gz · Last modified: 2025-02-13 10:25 by Merja Herzig

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