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Exponential Decay Exponential Radioactive Decay

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8/14/2013 Exponential Applications 2 Exponential Decay Radioactive Decay Radioactive isotopes of some elements such as 14 C, 16 N, 238 U, etc., decay spontaneously into more stable forms ( 12 C, 14 N, 236 U, 232 U, etc.) Decay times range from a few microseconds to thousands of years Decay rate proportional to amount present Same proportion decays in equal time

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8/14/2013 Exponential Applications 3 Exponential Decay Radioactive Decay Decay measurement Might not measure whole decay time Can measure limited decay, then calculate half-life Half-life = time for decay to half original measured amount Model with exponential function

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8/14/2013 Exponential Applications 4 Exponential Decay Radioactive Decay (continued) C A A A(3k) = (½)A(2k), …, A(nk) = A 0 (½) n, … = ( A 0 (½) ) (½) = A 0 (½) 2 = (A 0 (½) 2 )(½) = A 0 (½) 3 A(4k) = A 0 (½) 4 A(2k) = (½)A(k) Initial amount of radioactive substance Q is A 0 and decays to A(x) after x years After half-life of k years, A(k) = ½A 0 … or just A 0 (½) After 2k years amount is

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8/14/2013 Exponential Applications 5 Exponential Decay Radioactive Decay (continued) A, …, A(nk) = A 0 ( ½ ) n, … A(4k) = A 0 ( ½ ) 4 After n half-lives, x = nk so the amount A(x) = A(nk) = A 0 ( ½ ) n Since x = nk, then n = x/k and A(x) = A 0 ( ½ ) x/k left is

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8/14/2013 Exponential Applications 6 Exponential Decay Radioactive Decay (continued) A life can be modeled by an exponential function: Initial amount is f(0) = C and, for half-life k, = C((½) 1/k ) x = C(½) x/k a = (½) 1/k f(x) = C a x f(k) = (½)C = C a k Dividing out constant C gives: a k = ½ Solving for a, we get and thus f(x) = C a x Question: What does this look like graphically ? Alternate View: just recognize that half-

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8/14/2013 Exponential Applications 7 Exponential Decay Radioactive Decay Graph A0A0 A0A0 1 2 A0A0 1 4 A(t) t k2k3k 4k5k6k 0 A(t) = amount at A 0 = initial amount k = half-life A(t) = A 0 (½) t/k Since n = t/k A(t) = A 0 (½) n = A(0) A(t) reduced by half in each half-life After n half-lives t = nk A(t) = A 0 (½) t/k A0A0 1 8 A0A0 1 16 A0A0 1 32 time t

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8/14/2013 Exponential Applications 8 Think about it !

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8/14/2013 Exponential Applications 9 Solving Equations: Examples 1. World population 195019601970198019902000 20102020203020402050 15 14 13 12 11 10 9 8 7 6 5 4 3 2 t P(t) 1950 2.5098 1960 3.0000 1970 3.5859 1980 4.2862 1990 5.1233 2000 6.1239 2010 7.3199 2020 8.7495 2030 10.458 2040 12.500 2050 14.942 t P( t) P1P1 P2P2 P3P3 ( x 10 9 ) P(t) = 3(1.018) t–1960

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8/14/2013 Exponential Applications 10 Spare Parts Slide 2 2 | | | ± ½ ½ ½ ¼ ¼ ¼

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