Chem 312 Quiz 1
4 October 2006
Take home quiz
Name: ___________________
Question 1 (10 Points)
Use the chart of the nuclides to provide 7 facts about the isotope 154Eu.
1. spin 3
2. – parity
3. ground state half life 13.54 years
4. metastable state half life 46.1 min
5. decays by electron capture, position decay, and beta decay
6. beta decay energy 0.58 MeV
7. gamma decay energy 0.123 keV
8. gamma emission with decay
9. 91 neutrons
10. 63 protons
11. beta decay product is 154Gd
12. n, cross section 1300 barns
Question 2 (15 Points)
Define the following terms and provide 5 examples for each term (i.e., isotopes of the same
element, isotopes in the same isobar)
1. Isotopes: Same Z, different A
2. Isobars: Different Z, same A
3. Isotones: Same number of neutrons, different number of protons
4. Isomers: Long lived excited states of an isotope
Isomer: 98mNb, 99mNb, 99mTc, 60mCo, 182mTa
Isotopes: 98mNb, 99mNb, 99Nb
90
Mo, 99Mo
99m
Tc, 100Tc, 105Tc
104
Ru, 102Ru, 101Ru
Isotone: 98mNb, 99Mo, 100Tc, 101Ru, 102Rh
102
Ru, 99m, 99Nb
Isobar: 99Mo, 99mNb, 99Nb, 99mTc
102
Rh, 102Ru
1
Question 3 (15 Points)
1. Define mass excess
2. Using the mass excess data in the CD or the data found at
http://www.nndc.bnl.gov/wallet/wccurrent.html calculate the Q value for the following
reactions. Please show all data and work
a. Alpha decay of 244Pu
244
Pu+240U+Q
Q=244Pu-(240U+)
Q= 59.8056-( 52.7151+ 2.4249)
=4.666 MeV
b. The removal of a proton from 147Pm
147
Pmp+146Nd+Q
Q=147Pm-(146Nd+1H)
Q= -79.0479-( -80.9310+ 7.289)
Q=-5.4059 MeV
c. The removal of a neutron from 147Pm
147
146
Pmn+ Pm+Q
Q=147Pm-(146Pm+n)
Q=-79.0479-(-79.4599+8.0713)
Q=-7.6593 MeV
d. 248Cm(18O, 5n)261Rf
248
Cm+18O 5n+261Rf+Q
Q=248Cm+18O-(261Rf+5n)
Q= 67.3922+ -0.7815)-( 101.3154+5*8.0713)
Q=-75.061 MeV
Question 4 (10 Points)
Describe the liquid drop model and relate the terms of the equation to physical properties of the
nucleus.
2
2


 NZ 
 NZ 
2/3
2 1 / 3
2 1
E B  c1A 1  k
   c 2 A 1  k
   c3 Z A  c 4 Z A  
A
A

 

 


Nucleons can interact with only a small number of local nucleons. For this reason nucleons near
the surface will have different interactions than those near the center.
1st Term: Volume Energy in first approximation of the binding energy and is proportional to the
number of nucleons
2nd Term: Surface Energy: Nucleons at surface of nucleus have unsaturated forces and the term
decreasing importance with increasing nuclear size
3rd term represents the electrostatic energy that arises from the Coulomb repulsion between the
proton and lowers binding energy
4th term represents correction term for charge distribution with diffuse boundary
 term is the Pairing Energy
2
Question 5 (10 Points)
Define de and Re in the figure below.
Describe how nuclear dimensions are experimentally determined.
The half potential radius (Re) is about 6.5 fm and the skin thickness (de) is 2.25 fm.
Charge particles are projected towards a nucleus and scattered. Any positively charged
particle subject to nuclear forces can be used to probe the distance from the center of a nucleus
within which the nuclear (attractive) forces become significant relative to the Coulombic
(repulsive force). Positive and negative particles are used. Using moderate energies of electrons,
data is compatible with nuclei being spheres of uniformly distributed charges. High energy
electrons yield more detailed information about the charge distribution (no longer uniformly
charged spheres).
Question 6 (10 Points)
What is the activity (in Bq) of 100 µL of 1E-6 M 60Co?
Use A=N
For 
t1/2 = 5.272 years=1.66E8 s
=4.18E-9 s-1
For N
1E-6 mol/L * 100E-6 L * 6.02E23 atoms/mol = 6.02E13
A=6.02E13*4.18E-9 = 2.52E5 Bq
3
If this sample is counted for 1 minute on a detector with 15 % efficiency, estimate the relative
error in counts
2.52E5*.15*60=2.27E6 counts
relative error = sqrt(2.27E6)-1= 6.64E-4
Question 7 (15 Points)
You have discovered a new isotope. You detect three decay modes; alpha decay, beta decay and
spontaneous fission. You find that 65 % of the decays are from alpha decay and 10% of the
decays are spontaneous fission. The half-life is 75 seconds. What is the half-life each decay
mode?
t=++ SF
1=t +t + SF/t
t=ln2/75 = 9.24E-3 s-1
0.65=t
0.65*9.24E-3 s-1= 6.01E-3 s-1
alpha t1/2=ln2/6.01E-3 s-1=115 seconds
0.1= SF/t
0.1*9.24E-3=9.24E-4 s-1
SF t1/2 = ln2/9.24E-4 s-1=750 seconds
0.25=t
0.25*9.24E-3=2.31E-3 s-1
beta t1/2=ln2/2.31E-3 = 300 seconds
4
Question 8 (15 Points)
You place 10 g of KCl in a reactor with a thermal flux of 5E14 neutron cm-2sec-1 for 3 hours.
How many moles of 36Cl are produced?
Use the equation
The reaction under examination is:
35
Cl(n,)36Cl
The equation for the production of the isotope is:
R (atom/s)=N
From chart of the nuclides = 43.6E-24 cm2
=5E14 neutron cm-2sec-1
Need to find N (number of 35Cl atoms)
MW KCl= 74.551
10/74.511=0.134 moles Cl
75.77 % of Cl is 35Cl
0.102 moles 35Cl
35
Cl N = 0.102 moles x 6.02E23 atoms/mole= 6.14E22
R (atom/s)= 6.14E22(43.6E-24 cm2)(5E14 neutron cm-2sec-1)
R (atom/s)=1.34E15 atoms 36Cl/s
3 hours = 1.08E4 s
Total 1.45E19 atoms 36Cl
=1.45E19/6.02E23 = 2.40E-5 moles
5