After one unit of time, the number (P 1 ) of parents remaining is: P n = P 0. After two units of time, the number (P 2 ) of parents remaining is:

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1 Algebraic Equations for Radioactive Decay We begin with the experimental observation that the number of radioactive atoms of one isotope that decay in one unit of time is directly proportional to the number of radioactive atoms present. This means that the fraction (k) of radioactive atoms of one isotope remaining after the passage of one unit of time of length tk (e.g. one hundred years) will be a unique value (k) that depends only on the isotope ((such as 40 K) and the length of the unit of time (tk). In our analysis, this observation is expressed as a recursive formula, generalized, and then solved for the total time (t) that has passed since the start of the decay process. Let P 0 be the initial number of radioactive atoms ("parents") in a sample. Let k be the fraction of radioactive atoms remaining after one unit of time. Let tk be the length of the one unit of time. Let P n be the number of radioactive atoms ("parents") remaining in the sample after n units of time, where n is an integer. After one unit of time, the number (P 1 ) of parents remaining is: P 1 = P 0. k. (1) After two units of time, the number (P 2 ) of parents remaining is: P 2 = P 1. k. (2) Substituting for P 1 from the first equation (1) we have: P 2 = (P 0. k). k =. k 2. (3) After three units of time, the number (P 3 ) of parents remaining is: P 3 = P 2. k, (4) which becomes upon substitution for P 2 from equation (3): P 3 = (P 0. k 2 ). k = P 0. k 3. (5) By continuing this procedure, we can show that after n units of time (tk) the number of parents remaining is: P n = P 0. k n. (6) This equation expresses the fact that at any time the number of parents remaining is related in a simple way to the initial number of parent atoms. For example, if we select our unit of time tk so that k = (1/2), then one-half of the parents will remain after one unit of time. This unit of time, t½, which is different for each radioactive isotope, is

2 called the half-life of an isotope. Using (6), we can see that after 4 half-lives (i.e. 4 units of time), the number of parents remaining is: # 1 P 4 = P 0 % & $ 2' 4 ( # 1 & = P 0 % ( $ 16'. (7) To see what these equations mean, consider a specific example starting with P 0 = 64 radioactive atoms (e.g. 40 K). Plotting the number of radioactive atoms remaining and the number of daughter atoms accumulated as a function of the number of half-lives elapsed, we have: Parents [P(n)] Parents [P(n)] Daughters [D(n)] Daughters [D(n)] Half-Lives (n) As geologists, we are interested in using these relationships to find the total time (t) since the start of a decay process. To do so we start with a simple rearrangement of equation (6): " % Pn n $ ' = k. (8) # & Taking the logarithm of both sides of the equation and using the identity ln (a b ) = b ln(a) we have: " % Pn ln$ ' = n ln(k). (9) # & The total time t (yr) is related to the number of units of time n and the length of one unit of time tk (yr) as follows: t = n t k. (10)

3 Substituting for n in (9) using the relationship (10) and rearranging: " ln$ # Pn % ' = & ( t + ( ln(k) + * - ln(k) = t * -. (11) ), ), t k t k It has been shown experimentally that, for any one radioactive isotope, the term in brackets on the right of (11) has the same value whatever unit of time (tk) is selected. This value is used to define a "decay constant" (λ): $ λ - ln(k) ' & ), (12) % t k ( where λ is a positive number that has units of (yr) -1. The minus sign is used because k is a fraction and, therefore, ln(k) is negative. Using (12) to substitute for the term in brackets in (11) and changing notation ( = P n, where is the number parent atoms remaining at time t) we have: " ln$ # Pt % ' = t [-λ]. (13) & Rearranging (13) to solve for t, we obtain: t = - 1 λ ln # P & t % ( = 1 $ ' λ ln # P & 0 % (, (14) $ ' Pt which is the result we have been seeking. In these expressions, P 0 is the number parent atoms at time t=0, is the number parent atoms remaining at time t (yr), and λ (yr -1 ) is the decay constant. P o -- = P o exp (- λt) t 1/2 t = (-1/ λ) ln( /P o ) = (1/ λ) ln(p o / ) = (1/ λ) ln(2) t /(Half-Lives)

4 Our derivation of (14) assumes that t/tk is an integer n. This will not be correct for most times (t) if the value of k is a large fraction (e.g. ½) or randomly chosen. However, because the unit of time tk is of arbitrary size, we can always choose the value tk so that t/tk is an integer. Therefore, equation (14) is true for all cases and, indeed, is the same as the exact solution obtained using calculus to solve the differential equation dp dt = - λp (15) that expresses, for a continuous dynamic system, the experimental observation we used to obtain (1). Using (14) to find the time since the start of radioactive decay, all we need to do is discover the ratio of the number of parent atoms remaining ( ) to the initial number of parent atoms (P 0 ), once we know the value of the decay constant λ. We can do this for some cases by recalling that there will be one daughter atom produced for each parent that decays. The number of parent atoms we had to start with is given by the sum of the number of parent atoms left ( ) and the number of daughter atoms produced (D t ). If there were no daughter atoms present to start with, then all the daughter atoms in the rock must have been produced by radioactive decay. If this is the case our expression for t becomes: t = 1 ln " $ % λ # & = 1 ln " $ + D t% λ # & = 1 ln " $ λ 1 + Dt% # & (16) This is the expression that is used for 40 K/ 40 Ar dating of micas and amphiboles for which there is little likelihood of any initial 40 Ar being included in their crystal structures. If there were some daughter atoms present in the rock at the start of the radioactive decay, we have: P 0 = + (D t - D 0 ). (17) Since D 0 is just as much of an unknown as P 0, we cannot solve for t in cases where there were daughter atoms present at the start using this procedure. We still have a way, however, if we have more than one rock or mineral of the same age. To see how to do it we must rearrange (14) = P 0 exp(-λt). (18) Let us rewrite this equation as follows: P 0 = exp(λt) (19)

5 substituting for P 0 from (19) and rearranging D t = [exp(λt) - 1] + D 0. (20) Equation (5) is a very useful one. It is the equation of a line. If we had several samples of the same age that all started out with the same D 0, these samples would define a line on a graph of D t vs.. The slope of this line would be: [exp (λt) -1] (λt) if (λt) is small. Thus, to find the time since the start of the decay process, we need only graph the data and measure the slope of the line defined by (20). The catch is finding several samples of the same age that had the same D 0. In fact, this is probably impossible to do. However, by a little chicanery we can still get the desired result. We can get several samples of the same age by considering several different minerals from the same igneous rock. But because different minerals have different crystal structures and chemical composition, there is no guarantee that they all had the same D 0. To understand the "trick" we need to use let us consider a specific example. Let the parent be 87 Rb and the daughter be 87 Sr. Equation (20) becomes for this example: ( 87 Sr) t = λt ( 87 Rb) t + ( 87 Sr) 0. (21) Now it happens that there is another common isotope of strontium, 86 Sr. 86 Sr is not produced by any radioactive decay process. Thus, the amount of 86 Sr in a sample does not change with time. In other words ( 86 Sr) t = ( 86 Sr) 0. (22) Let us divide the expression (6) by ( 86 Sr) t ( 87 Sr) t 86 Sr ( ) t = λt ( 87 Rb) t 86 Sr ( ) t + ( 87 Sr) 0 86 Sr ( ) 0. (23) Once again in (23) we have the equation of a line. If we have several samples of the same age with the same initial ( 87 Sr)/( 86 Sr) ratio, a graph of ( 87 Sr) t /( 86 Sr) t vs. ( 87 Rb) t /( 86 Sr) t for these samples will define a line whose slope is given by (λt). In this case, however, it is quite reasonable to assume that all the minerals of an igneous rock had the same ( 87 Sr) 0 /( 86 Sr) 0 ratio. The reason is that simple chemical processes will not distinguish between these two isotopes. No mineral has a preference for 87 Sr over 86 Sr because they have the same ionic radius and charge. Thus, all minerals that crystallize from the same magma will have the ( 87 Sr) 0 /( 86 Sr) 0 ratio of that magma!

6 The equations presented here are based on three important assumptions: 1. The number of radioactive atoms of a given type decaying in one unit of time is proportional to the number of radioactive atoms present. This assumption has been verified by observation for short-lived radioactive isotopes. 2. The decay constant (λ) has not changed during geologic time and remains constant regardless of local physical or chemical changes. 3. The samples we wish to date represent a closed chemical system from the time of their formation to the present. In other words, there has been no addition or loss of parent atoms or daughter atoms except through the process of radioactive decay. Assumptions 1 and 2 above are general assumptions for radiometric dating. The correctness of assumption 3 depends directly on the geologic history of the sample in question. For example, metamorphism may heat minerals to temperatures high enough for loss of daughter atoms like 40 Ar. Therefore every dated sample must be judged on a case-by-case basis to determine if assumption 3 is valid. If we have good reason to believe that D 0 = 0, then equation (16) may be used directly to obtain the desired age t. Usually D 0 = 0 holds for samples of micas and amphiboles to be dated using the K/Ar method. Why is this so? We know that D 0 0 for samples to be dated using the very versatile 87 Rb/ 87 Sr dating method because initial 87 Sr occurs in most datable rock systems. In this case we must determine graphically the initial ratio ( 87 Sr) 0 /( 86 Sr) 0. Decay Parent Daughter Constant Half-life Isotope Isotope (per year) (billion years) Decay Mechanism(s) 235U 207Pb x alpha and beta decay in series 238U 206Pb x alpha and beta decay in series 40K 40Ar 5.81x electron capture 232Th 208Pb 4.948x alpha and beta decay in series 176Lu 176Hf 1.93x beta decay 187Re 187Os 1.612x beta decay 87Rb 87Sr 1.42x beta decay 147Sm 143Nd 6.54x alpha decay (data from Dalrymple, 1991, p. 80; Faure, 1986)