=ln2/t 1/2. Find the half life in seconds and calculate. Quiz 2 4 October 2012

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1 CHEM 312 Quiz 2 4 October 2012 Name: Assigned: 4 October 2012 Due: 16 October 2012 The test is take-home. Please feel free to contact me with questions. Lecture 1 (Introduction) after slide 1-31, Lecture 2 Nuclear Properties, Lecture 3 Decay Kinetics, Chapter 4 Alpha Decay Use the chart of the nuclides, table of the isotopes, and/or data from the web links to answer the following questions. Use the chart of the nuclides, table of the isotopes, the mass excess data in the textbooks, and/or data from the web links from lectures 1 to 4 (or any you find) to answer the following questions. 1. (10 Points) Provide the decay constants (in s -1 ) for the following isotopes Isotope (s -1 ) 50 Ca 81 Zn 95 Zr 137 Cs 151 Sm 241 Am 242 Pu =ln2/t 1/2. Find the half life in seconds and calculate. Isotope t 1/2 s (s -1 ) 50 Ca 1.40E E Zr 5.53E E Sm 2.84E E Am 1.37E E Pu 1.18E E-14 1

2 2. (15 Points) You have a sample of 95 Zr that is counted for 1 minute and has an activity of 312 Bq. Consider this the activity at time zero. Please provide the activity in Bq and the % error in the value for the following times after time zero for 1 minute counting. Assume you have 100 % detection of each decay, so the total counts is equal to the total decay. Time (days) Activity (Bq) % error Use A t =Ae -t to find the activity at the designated time. The decay constant is ln2/64.03 d -1 The activity is in Bq. The counts found in 1 minute are A(Bq) 60 = counts at time t (C t ). The absolute error is the square root of the counts. The relative error is the inverse. The error is based on the total counts (A in Bq for 60 seconds) time (days) A (Bq) Counts absolute error relative error % error E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E+00 2

3 3. (5 Points) How long would it take to decay 95 % of 99 Mo? Use N/N 0 =e -t If 95 % decays, 5 % remains, therefore N/N 0 = 0.05 The half life of 99 Mo is hours, so = hr -1. Re-arrange the equation and solve for t t=ln(n/n o )/ - t=ln0.05/ hr -1 = 285 hours = 11.9 days 4. (5 Points) List the 4 stable odd-odd isotopes The first 4 odd Z isotopes are the examples. 2 H, 6 Li, 10 B, 14 N 5. (10 Points) Calculate the Q value for the following reactions Reaction or decay Alpha decay of 239 Pu Q value (MeV) 45 Sc + 4 He 5 H+ 44 Ti decay of 137 Cs n+ 235 U 232 Th+α n+ 235 U 236 U From the Q value calculator or data in Table of the isotopes (for 137 Cs decay) Reaction or decay Q value (MeV) Alpha decay of 239 Pu Sc + 4 He 5 H+ 44 Ti Beta decay of 137 Cs n+ 235 U 232 Th+α n+ 235 U 236 U

4 6. (10 points)consider the irradiation of 1 g of Mn metal in a thermal neutron flux of 2.5x10 14 n/cm 2 /sec. How many atoms of 56 Mn are produced after the following irradiation times The half life of 56 Mn is hours. The irradiation time is on the order of the half life. Decay of the produced 56 Mn will occur during the irradiation. Use the equation that includes the saturation factor 1t ( 1 e ) N 1 N 0 1t ( 1 e ) 1 The reaction is 55 Mn(n,) 56 Mn =13.3E-24 cm 2 N o = 1 g Mn x 1 mole/ g x 6.02E23 atoms 55 Mn/mole Mn = E+22 atoms 55 Mn 56 Mn t ½ =2.578 hours= seconds. The time should be in seconds as the flux is in seconds. =07.47E-5 s -1 =2.5x10 14 n/cm 2 /s Put the data into the equation, set t for the different times and solve for N 1. Since is in s -1, the time t should be in seconds. No sigma flux time sec N1 (atoms 56 Mn produced) 1.10E E E E E E E E+17 a. 1.5 hours N 1 = E+17 atoms of 56 Mn b. 4.5 hours N 1 = E+17 atoms of 56 Mn 4

5 7. (10 Points) Using 14 C dating, what is the age of a wooden sample with 0.10 Bq/g C? From the lecture 1 t ln( C C eq sample ) 14 C eq is 227 Bq 14 C /kgc, or Bq 14 C /gc =ln(2)/5730 =1.21E-4 y -1 Substituting into the equation t ln( ) years 1.21E The object is 6775 years old. 5

6 8. (15 Points) Calculate the alpha decay Q value and Coulomb barrier potential for the following. Which isotope has the largest difference between Q value and Coulomb barrier for alpha decay? The Q-value is determined from the Q-value calculator web site. The coulomb barrier is 2( Z) c 1. 44MeV 1/ 3 1/ 1.2( A 4 ) V 3 Where Z and A are for the daughter. The other Z=2 and A=4 is for the alpha particle. a. 212 Bi b. 210 Po c. 238 Pu d. 239 Pu e. 240 Am f. 241 Am Isotope Z A Q value (MeV) Vc (MeV) delta V-Q 212 Bi Po Pu Pu Am Am Largest difference with 239 Pu 6

7 9. (20 points) Use the following even Pu isotope data to answer the following. Decay data from ( Pu isotope Isotope T 1/2 Q (alpha) kev % alpha branch of decay min hr y y E+5 y E d a. Calculate and show the resulting Geiger Nuttall relationship for the even A Pu isotopes. The general equation is Q is the alpha decay Q value. The half life is in seconds. The Geiger-Nuttall relationship is log t 1/2 = A+B/(Q ) 0.5 From the data above plot the log of the alpha half-life against the inverse of the square root of the Q value. The slope is B and the intercept is A. The alpha decay half life can be found by the relationship % alpha branch = t The value for t = ln(2)/t 1/2 (put half life in seconds) When is found then t 1/2 alpha is found. Plot log t 1/2 (alpha) against square root of Q, solve for A and B Pu isotope The data are below. The decay constants are in sec -1 and the Q value is in kev. Isotope T 1/2 Q (alpha) kev % alpha branch of decay T 1/2 sec tot t 1/2 Log t 1/2 1/sqrt Q min E E E E E hr E E E E E y E E E E E y E E E E E y E E E E E E+5 y E E E E E E E E E E E d

8 1/Q sqrt y = m1 + m2 * M0 Value Error m m Chisq NA R NA The resulting equation is log t 1/2 alpha b. Use the relationship to estimate the alpha decay half life for 246 Pu. To determine the alpha decay half life for 246 Pu use the equation above and the Q value. The Q value can be determined from ( The Q value is KeV. Inserting this value into the equation above yields Log t 1/2 =18.0 T 1/2 = second = 1.00E18 seconds=3.17e10 years c. Use the relationship to determine the hindrance factors for the odd A Pu isotopes listed in the table below. Calculate the hindrance factor from the ratio of the measures alpha decay half life over the calculate half life Isotope % alpha branch of decay Isotope T 1/2 Mass Excess Pu Isotope A (MeV) Mass Excess U Isotope (A-4) (MeV) Hindrance factor min d y y Mass Excess alpha=2.425 MeV. Mass excess data from Nuclear and Radiochemistry, 3 rd Edition 8

9 The hindrance factor is the ratio of measure alpha decay half life to the calculated. The measured alpha decay half life is from the % alpha branch and isotope half life. % alpha branch = t The value for t = ln(2)/t 1/2 (put half life in seconds) When is found then t 1/2 alpha is found. This is the measured half life. Calculating the Q value for the alpha decay from the data in the table provides the calculated half life using the equation: The ratio of the measured and calculated provides the Hindrance Factor. For the Q value the alpha mass excess is MeV. Q values need to be in kev for equation above. Isotope % alpha branch of decay Isotope T 1/2 Mass Excess Pu Isotope A (MeV) Mass Excess U Isotope (A- 4) (MeV) t 1/2 al measured sec Q alpha (kev) T 1/2 calc sec Hindrance factor min E E E E d E E E E y E E E E y E E E E+01 9