The below parameters are provided for convenience. It does not imply you have to
ID: 107621 • Letter: T
Question
The below parameters are provided for convenience. It does not imply you have to use all of them in your problem solving
Samples of gold-bearing conglomerate from the Witwatersrand mining district of South Africa were analyzed byNicholaysen et al. (1962, Geochim. Cosmochm, Acta, 26:15-23). 20 Pb 207 Pb 208 Pb Sample U Pb 20+ 20+ Pb 20+ B 153 2.46 1.77 571 142 52.6 KCGI 0.201 0.112 249 68.3 62.6 KCGIV 0.520 0.350 326 84.7 69.1 The initial isotope ratios are 204:206:207 208 1.0:12.4:14.4:32.7. Calculate the age of these samples by means of a U-Pb Concordia diagramExplanation / Answer
Radioactive decay is described in terms of the probability that a constituent particle of the nucleus of an atom will escape through the potential (Energy) barrier which bonds them to the nucleus. The energies involved are so large, and the nucleus is so small that physical conditions in the Earth (i.e. T and P) cannot affect the rate of decay. The rate of decay or rate of change of the number N of particles is proportional to the number present at any time, i.e. dN/dt is inversely proportional to N.
Note that dN/dt must be negative.
The proportionality constant is lambda, the decay constant. So, we can write dN/dt = - lambda N
Rearranging, and integrating, we get
ln(N/N0) = -l(t – t0)
If we let to = 0, i.e. the time the process started, then N = N0e-lamda*t
We next define the half-life, t1/2, the time necessary for 1/2 of the atoms present to decay.
This is where N = No/2. Thus, N0/2 = N0e-lamda*t so that t1/2 = ln*2 / lambda
By definition, D* = N0 – N , N = N0e-lambda*t
D* = Nelamda*t – N = N(elambda*t – 1)
Now we can calculate the age if we know the number of daughter atoms produced by decay, D* and the number of parent atoms now present, N. The only problem is that we only know the number of daughter atoms now present, and some of those may have been present prior to the start of our clock. By Beta decay the neutron emits an electron to become a proton. For this decay reaction,
Lambda = 1.42* 10-11 / yr , t1/2 = 4.8 * 1010 yr
To account for this, we first note that there is an isotope of Sr, 86Sr, that is:
(1) non-radiogenic (not produced by another radioactive decay process),
(2) non-radioactive (does not decay to anything else).
Note also that equation has the form of a linear equation, i.e. y = mx +b
Two isotopes of Uranium and one isotope of Th are radioactive and decay to produce various isotopes of Pb. The decay schemes are as follows:
23U = 8 4He + 206Pb by alpha decay
Lambda238 = 1.551*10-10/ yr, t1/2 = 0.707 * 109 yr
Note that the present ratio of 235U / 238U = 1 / 137.8
204Pb is a stable non-radiogenic isotope of Pb, so we can write two isochron equations and get two independent dates from the U - Pb system.
(206Pb / 204PB)t = (206Pb / 204PB)0 + (238U/204Pb)t(elambda238t – 1)
As per the Concordia diagram the values would represent as follows:
Time
Age of 206Pb / 238U
0
0
0.2
0.5
0.4
1.0
0.5
1.5
0.6
2.0
0.7
2.5
0.8
3.0
0.81
3.5
0.9
4.0
0.92
4.5
Time
Age of 206Pb / 238U
0
0
0.2
0.5
0.4
1.0
0.5
1.5
0.6
2.0
0.7
2.5
0.8
3.0
0.81
3.5
0.9
4.0
0.92
4.5
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