Chem 312
Homework 6: Lecture 6 (Alpha Decay)
Assigned Wednesday 11 October, Due Monday 16 October, Homework Quiz 6 Wednesday 18 October.
1. Determine if the following isotopes can decay by alpha emission. Use mass excess data.
a. 145Eu
145Euà141Pm+a+Q
Q=-77.998-(-80.48+2.425)
Q=0.057 MeV
Q is positive so alpha decay is possible
b. 159Ho
159Hoà155Tb+a+Q
Q=-67.342-(-71.26+2.425)
Q=1.493 MeV
Q is positive so alpha decay is possible
c. 120Xe
120Xeà116Te+a+Q
Q=-82.030-(-85.28+2.425)
Q=0.825 MeV
Q is positive so alpha decay is possible
d. 190Pt
190Ptà186Os+a+Q
Q=-37.338-(-43.007+2.425)
Q=3.244 MeV
Q is positive so alpha decay is possible
e. 103Pd
103Pdà99Ru+a+Q
Q=-87.455-(-87.618+2.425)
Q=-2.262 MeV
Alpha decay is not energically favored
f. 203Hg
203Hgà199Pt+a+Q
Q=-25.292-(-27.430+2.425)
Q=-0.287 MeV
Alpha decay is not energically favored
2. Using the results from question 1, identify those isotopes that can decay by alpha but the alpha decay is not observed. Explain this observation.
The isotopes that can decay by alpha emission based on energetics are:
145Eu, 159Ho, 120Xe, and 190Pt
Of these isotopes only 190Pt has an observed alpha decay with a half-life of 7E11 years. The Q values for the alpha decay of the other isotopes are all below 1.5 MeV, not even half of the value of 190Pt. The alpha decay half lives of these isotopes would be expected to be extremely long. The isotopes 145Eu, 159Ho, and 120Xe are neutron light and decay by EC or positron emission and have half-lives of 5.93d, 33 m and 40 m respectively. The long-half life of the alpha decay branch does not meaningfully compete with the short overall isotopic half-life so the decay is not observed.
3. Show the equations that relate the Q value to the alpha decay energy.
• Conservation of momentum in decay, daughter and alpha are equal rd=ra ,
p2=2mT where m= mass and T=kinetic energy for a and daughter so 2maTa=2mdTd
Td= maTa/md
4. Use the relationship from question 3 to determine the alpha decay energy for 145Eu, 159Ho, 120Xe, and 190Pt
145Eu, Q=0.057 MeV
Ta =0.057(141/145)=0.055 MeV
159Ho, Q=1.493 MeV
Ta =1.493(155/159)=1.455 MeV
120Xe, Q=0.825 MeV
Ta =0.825 (116/120)=0.7975 MeV
190Pt, Q=3.244 MeV
Ta =3.244 (186/190)=3.176 MeV
5. What is the theory for alpha decay?
• Distance of closest approach for scattering of a 4.2 MeV alpha particle is ~62 fm.
o Distance at which the alpha particle stops moving towards the daughter
o Repulsion from Coulomb barrier
• An alpha particle should not get near the nucleus or
• For decay
o alpha particle should be trapped behind a potential energy barrier
6. What are some general trends observed in alpha decay?
• For the isotopes of any element in the region of the a emitters the a-decay energies are expected to decrease monotonically with increasing A
• For a series of isobars energies will increase with increasing Z
• Data shows exceptionally large binding energies just below neutron number 126
– evidence for closed shell at N=126
• Considering only binding-energy, spontaneous emission of tightly bound nuclear entities other than a particles would be permitted
– barrier heights for the emission of such nuclei, though, are so high that the expected half lives are beyond presently detectable limits
• Correlation between half-life and decay energy for similar isotopes (even or odd) for a given element
7. Use the following data to calculate the half-life and Q value for the alpha decay of 228U with Qa values are 6.803 MeV.
Isotope / t1/2 (years) / Qa (MeV)232U / 69.8 / 5.413
233U / 1.59E5 / 4.908
234U / 2.46E5 / 4.857
235U / 7.04E8 / 4.678
236U / 2.34E7 / 4.573
238U / 4.47E9 / 4.269
The relationship between half-life and Q value is:
Log t1/2=A+BQa0.5
Plot log t1/2 against Qa only for the even isotopes. The alpha decay hindrance precludes the inclusion of the odd isotopes.
The values for the equation above are:
Log t1/2=71.37-29.91Qa0.5
isotope / Q / log t1/2 / t 1/2 a / t 1/2 s228U / 6.803 / -6.64294 / 2.25E-07 / 7.18
The actual value is 9.1 m