ES 6216 – Isotope Geochemistry
Solution Set 1
1. A hypothetical water sample is characterized by the following mole fractions:
n 17O
= 0.00039
n 16O
n 18O
= 0.00210
n 16O
a) Calculate the isotopic abundance of each of the three stable isotopes.
Isotopic abundances are expressed as percentages:
18
Omoles
abun 18O =
*100
total (Omoles)
So for every 1 mole of 16O, you have:
n 17O
*1mol 16O = 0.00039*1 = 0.00039mol 17O
16
n O
n 18O
*1mol 16O = 0.00210*1 = 0.0021mol 18O
16
n O
Total O moles = 1mol 16O + 0.00039mol 17O + 0.0021mol 18O = 1.00249 mol O
And abundances are as follows:
0.00039mol
abun 17O =
*100 = 0.0389%
1.00249mol
0.0021mol
abun 18O =
*100 = 0.2095%
1.00249mol
1.0mol
abun 16O =
*100 = 99.75%
1.00249mol
b) Calculate the atomic weight of the oxygen in this sample.
Atomic wt 16O=15.994915 amu (from table of nuclides)
17
O=16.999132 amu
18
O=17.999160 amu
and using the abundances calculated in part a),
atomic wt. O in sample = (0.9975*15.994915amu)+(0.000389*16.999132amu)+
(0.002095*17.999160amu)=15.99922amu
2. Calculate the binding energy (in Mev) associated with 40Ca and 40K. [watch your units]
First we derive the energy associated with 1amu mass difference. And for that we
need to calculate the weight of 1amu in grams:
1amu =
1 12g
1mol
1
*
*
= grams = 1.6606e−24 g
12 mol 6.022e23atoms / mol N
Next we use E=mc2 to calculate binding energy:
where c=2.9975e10 cm/sec, and
E = (1.6606e −24 g ) *(2.9975e10cm / sec) 2 = 1.492e −3ergs
1.492e−3ergs
= 9.3128e8eV /(106 eV / MeV ) = 931.28MeV
−12
1.6021e erg / eV
1
Now we calculate the Δamu for 40Ca:
Published atomic weight = 39.962591amu
Nominal mass = 20 protons + 20 neutrons
= 20(1.08866491amu) + 20(1.00782503amu) = 40.3297988 amu
Δm = 0.03672082amu
ΔE = 0.03672082amu(931.28MeV/amu) = 341.44 MeV or 8.5493 MeV/nucleon
The same set of calculations for 40K (21 neutrons + 19 protons) yields ~ 8.5165
MeV/nucleon. This is lower than 40Ca, which it should be, because it’s radioactive.
3. Consider the fusion reaction: 11H + 12 H → 23 He + ΔE
a) Calculate ΔE (in Mev) for this reaction.
First we need the atomic masses of the nuclides involved:
1
1 H = 1.007825amu
2
1
H = 2.014102amu
3
2
He = 3.016049amu
Then calculate Δm, which is related to ΔE through the equation E=mc2
Δm=1.007825amu+2.014102amu-3.016049amu=0.005905amu
ΔE=0.005905amu(1.6606e-24g/amu)*(2.9975e10cm/sec)2
=8.8106e-6ergs/(1.6021e-12ergs/ev)/106=5.4994MeV
b) If the observed 3He flux from the Earth is 2x1019 atoms/s, what fraction of
Earth’s heat flux (1.2x10-6cal/cm2/s) can be explained by this reaction?
The Earth’s surface area is 5.1e18cm2, so the 3He flux is:
(2e19atoms/sec)/5.1e18cm2 = 3.92 atoms/cm2/sec
and the heat flux associated with this atom flux is:
3.92atoms/cm2/sec*5.4994MeV/atom=21.566MeV/cm2/sec
We need to convert the Earths’ heat flux into MeV to calculate a percentage:
1.2e-6cal/cm2/sec* (4.184 J/cal)*(107 ergs/J)/(1.6021e-12ergs/eV)/106 =
3.3134e7 MeV/cm2/sec
And the percentage contributed by 3He fusion is:
(21.566MeV/cm2/sec) / (3.3143e7 MeV/cm2/sec) *100 = 6.88 x 10-5
4. How much 14C would you expect to find in a 1g wood sample formed in 2000 B.C.?
Given: T=4012 y
A0=13.56 dpm/g C
t1/2=5730y
λ=0.693/t1/2=1.209e-4 yr-1
−
(1.209e −4 yr −1*4012 y )
) = 8.350dpm
A=A0e- T , so A = 13.56dpm(e
-4 -1
A=λN , so 8.350dpm=(1.209e yr /(365d/y)/(24hr/d)/(60min/hr))* N
so N=3.63 x 1010 atoms 14C
λ
5. Calculate the activity (in dpm/liter) of 40K and 238U in seawater. [K]=400ppm; [U]=3ppb
[Hint: you will first need to calculate the atoms/liter for K and U]
400ppm=400x10-6 g K per g of seawater = 10mmol K/kg s.w.
2
1liter s.w. = 1.025 kg, so we have 10.25mmol K/liter s.w.
10.25mmol/l * 6.022e23 atoms/mol = 6.15e21 atoms K/l
A similar calculation for U yields 7.38e15 atoms U/l
Next we need to calculate the # atoms/l of 40K and 238U, using the abundance table:
abun 40K = 0.0117%*(6.15e21 atoms K/l)= 7.19e17 atoms 40K/l
abun 238U = 99.27%*(7.38e15 atoms U/l)= 7.33e15 atoms 238U/l
Finally we calculate A we use the equation A=λN:
A(40K)=7.19e17 atoms/l * 5.543e-10 y-1=3.985e8 dpy/l= 758 dpm/l
A(238U)=7.33e15 atoms/l * 1.55e-10 y-1=1.136e6 dpy/l= 2.2 dpm/l
6. If the 238U series isotopes in seawater were in secular equilibrium (they are not), what
would be their combined activities (in dpm/liter)?
We just calculated that the activity of 238U in seawater is 2.2dpm/l, so if the 238U decay series
is in secular equilibrium, we can assign this activity to all the radioactive daughters in the
238
U decay chain (N=14):
Combined A = 14*2.2dpm/l=30.8dpm/l
7. You are charged with measuring the half-life of 238U. Like a smart geochemist, you take
a 3.5g rock that has a known 238U concentration of 1.8ppm, separate the 238U, and plate
it onto a alpha-counter source and measure its decay rate. How long would you have to
count to determine the half-life to a 1σ accuracy of 1%? What if you needed to know
the half-life to a 2σ accuracy of 0.1%?
Counting statistics dictate that to achieve a 1σ accuracy of 1%, we must accumulate 10,000
counts (0.01=1/sqrt(N)). After your geochemical separation is done, you have:
3.5g * 1.8e-6g 238U / g rock = 6.3e-6 g 238U = 2.647e-8 moles 238U= 1.59e16 atoms
The activity of these atoms can be calculated from A=λN,
A=(0.693/4.5e9y)*1.59e16 atoms = 2.45e6 dpy = 4.67 dpm
Assuming that the A doesn’t change appreciably over the relatively short
measurement window (you can convince yourself of this by going through the
A=A0e- T equation), then: 10,000 counts/4.67 dpm = 2141 minutes = 1.5 days
λ
b) for a 2σ accuracy of 0.1%, you’ll need N=(2/0.001)^2 counts = 4e6 counts
which will take 4e6 counts/ 4.67 dpm = 595 days
3