chapter4

# E the process of estimating the age of an artifact is

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e) The process of estimating the age of an artifact is called carbon dating . Example 4.6 : Suppose that we have an artifact, say a piece of fossilized wood, and measurements show that the ratio of C-14 to carbon in the sample is 37% of the current ratio. Let us assume that the wood died at time 0, then compute the time T it would take for one gram of the radio active carbon to decay this amount. Solution: By model (1.10) km dt dm = This is a separable differential equation. Write it in the form kdt dm m = 1 Integrate it to obtain ln|m|=kt+c Since mass is positive, lml=m and ln(m)=kt+c. Then 89

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m(t) = e kt+c =Ae kt , where A = e c is positive constant. Let at some time, designated at time zero, there are M grams present. This is called the initial mass. Then m(o) = A = M, so m(t) = Me kt. If at some later time T we find that there are M T grams, then m(T) = M T = Me kT . Then kT M M ln T = hence = M T M T k ln 1 This gives us k and determines the mass at any time: m(t) = M M ln T t T Me Let T= λ be the time at which half of the mass has radiated away, that is, half-life. At this time, half of the mass remains, so M T =M/2 and M T /M = 2 1 . Now the expression for mass becomes m(t) = ) 2 1 ln( t Me λ or m(t) = 2 ln t Me λ - Half-life of C-14 is 5600 years approximately, that is, 90
λ = 5600 00012378 . 0 5600 2 ln - means approximately equal (all decimal places are not listed). Therefore m(t)=Me -0.00012378t or t e M t m 00012378 . 0 37 . 0 ) ( - = = by the given condition that M t m ) ( is .37 during t. T= - = 00012378 . 0 ) 37 . 0 ln( 8031 years approximately. Example 4.7 (a) A fossilized bone is found to contain one thousandth the original amount of C-14. Determine the age of fossil. (b) Use the information provided in part (a) to determine the approximate age of a piece of wood found in an archaeological excavation at the site to date prehistoric paintings and drawing on the walls and ceilings of a cave in Lascaux, France, provided 85.5% of the C-14 had decayed. Solution: a. The separable differential equation ) ( t N k dt dN = , where k is the constant of proportionality of decay, models the phenomenon as discussed above. The solution is N(t) = N 0 e kt (say λ = -k, if we want to put in the form of the above discussion). 91

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Half-life of C-14 is approximately 5600 years 2 N o = N(5600) or 2 1 N 0 = N 0 e 5600k . By cancelling N 0 and taking logarithm of both sides we get 2 ln 2 1 ln 5600 - = = k or k= - 5600 2 ln = -0.00012378 Therefore N(t) = N 0 e -0.00012378t With N(t) = 0 1000 1 N we have 1000 1 N 0 =N 0 e -0.00012378t -0.00012378t = ln 1000 1 = - ln 1000. Thus yrs 55800 00012378 . 0 1000 ln t = (b) Let N(t)= N 0 e kt where k= -0.00012378 by part (a). 85.5% of C-14 had decayed; that is, N(t) = 0.145 N 0 or N 0 e - 0.00012378t = 0.145 No Taking logarithm of both sides and solving for t, we get t 15,600 years 4.4 Mixture of Two Salt Solutions 92
Example. 4.8 A tank contains 300 litres of fluid in which 20 grams of salt is dissolved. Brine containing 1 gm of salt per litre is then pumped into the tank at a rate of 4 L/min; the well-mixed solution is pumped out at the same rate. Find the number N(t) of grams of salt in the tank at time t.

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