chapter4

# 150 200 e 2k 85 or 4 3 2 k e or 2kln 4 3 this gives 2

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150 = 200 e -2k 85

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or 4 3 2 = - k e or –2k=ln 4 3 This gives 2 1 3 4 ln 2 1 k = = (0.2877)= 0.1438 0.14 The mass of the isotope remaining after t years is then given by m(t) =200e -.1438t The half-life t h is the time corresponding to m=100mg. Thus 100 = 200 e -0.14t h or 2 1 = e -0.14t h or t h = - years 95 . 4 14 . 0 693 . 0 5 . 0 ln 14 . 0 1 = - - = Carbon Dating: The key to the carbon dating of paintings and other materials such as fossils and rocks lies in the phenomenon of radioactivity discovered at the turn of the century. The physicist Rutherford and his colleagues showed that the atoms of certain radioactive elements are unstable and that within a given time period a fixed portion of the atoms spontaneously disintegrate to form atoms of a new element. Because radioactivity is a property of the atom, Rutherford showed that the radioactivity of a substance is directly proportional to the number of atoms of the substance present. Thus, if N(t) denotes the number of atoms 86
present at time t, then dt dN , the number of atoms that disintegrate per unit time, is proportional to N; that is, - = dt dN λ N (4.5) The constant λ , which is positive, is known as the decay constant of the substance. The larger λ is, the faster the substance decays. To compute the half life of substance in terms of λ , assume that at time t=t 0 , N(t 0 )=N 0 . The solution of the initial value problem - = dt dN λ N N(t 0 ) = N 0 (4.6) is N(t)=N 0 e - λ (t-t o ) or o N N e - λ (t-t o ) Taking logarithms of both sides we obtain - λ (t-t 0 )=ln o N N (4.7) If o N N = 2 1 , then - λ (t-t 0 )=ln 2 1 , so that t-t 0 = 6931 . 0 2 ln λ = λ Thus the half life of a substance is ln2 divided by the decay constant λ . 87

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The half-life of many substances have been determined and are well published. For example, half-life of carbon-14 is 5568 years, and the half-life of uranium 238 is 4.5 billion years. Remark 4.3.1 a) In (4.5) λ is positive and is decay constant. We may write equation (4.5) in the form = dt dN λ N, where λ is negative constant, that is, λ <0. b) The dimension of λ is reciprocal time. It t is measured in years, then λ has the dimension of reciprocal years, and if t is measured in minutes, then λ has the dimension of reciprocal minutes. c) From (4.7) we can solve for t-t 0 = N N ln 1 o λ (4.8) If t 0 is the time the substance was initially formed or manufactured, then the age of the substance is N N ln 1 0 λ . The decay constant λ is known or can be computed in most cases. N can be computed quite usually. Computation or pre- knowledge of N 0 will yield the age of the substance. By the Libby’s discovery discussed in Section 1.4.2. the present rate R(t) of disintegration of the C-14 in the sample is given by R(t)= λ N(t)= λ N 0 e - λ t and the original rate of disintegration is R(o)= λ N 0 . Thus t e R t R λ - = ) 0 ( ) ( so that 88
t= ) ( ) ( ln 1 t R o R λ (4.9) d) If we measure R(t), that present rate of disintegration of the C-14 in the charcoal and observe that R(o) must equal the rate of disintegration of the C-14 in the comparable amount of living wood then we can compute the age t of the charcoal.

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