Homework2Sol - Homework 2 AMATH 383, Autumn 2009 Due:...

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Homework 2 AMATH 383, Autumn 2009 Due: Monday, October 26, after class 1. (6 points) Age of Cave Painting: The caves near Lascaux, France, contain multiple wall paintings of animals from ancient times. To determine the age charcoal found in the caves have been investigated using the C-14 dating method. The charcoal gave an average count of R measured = 0 . 97 disintegrations min gram of C-14 in a sample. Fresh wood measurements today give R fresh = 6 . 68 disintegrations min gram . Estimate the age of the charcoal and hence the possible date of the wall paintings. Solution: C-14 concentration y ( t ) = y 0 e - λt with λ = 1 . 245 × 10 - 4 1 year R ( t ) = - d y d t = λy 0 e - λt R (0) = λy 0 ! = 6 . 68 R ( T ) = λy 0 e - λT ! = 0 . 97 T = - 1 λ ln R ( T ) R (0) = 15498years 2. (1+4+10+10+5+5+1 points) Reduced Thorium Chain: When restricted to the slowest compo- nents the Thorium chain reduces to Pu-244 τ 1 -→ U-236 τ 2 -→ Th-232 τ 3 -→ Pb-208 with half-lifes τ 1 , 2 , 3 around 10 8 years. In order to investigate this chain analytically we make the simplifying assumptions for the decay constants λ i = ln2 i λ 3 = 1 10 λ 1 and λ 2 = 10 λ 1 (a) Which decay is the fastest and which one the slowest in this chain? (b) Explain why the system y 0 1 ( t ) = - λ 1 y 1 ( t ) y 0 2 ( t ) = - λ 2 y 2 ( t ) + λ 1 y 1 ( t ) y 0 3 ( t ) = - λ 3 y 3 ( t ) + λ 2 y 2 ( t ) y 0 4 ( t ) = + λ 3 y 3 ( t ) describes the above chain. What elements are the function y 1 , 2 , 3 , 4 refering to? 1
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Homework, AMATH 383, Autumn 2009 (c) Write the system in matrix form. The general solution of the system is a superposition of solutions of the form y ( t ) = v exp( αt ) with α R , v R 4 and y ( t ) = ( y 1 ( t ) ,y 2 ( t ) ,y 3 ( t ) ,y 4 ( t )) T . What are the possible values for α and for the vectors v ? (d) We choose as initial condition the situation where only Pu-244 is given (891 units, say) and no other element, that is ( y 1 (0) ,y 2 (0) ,y 3 (0) ,y 4 (0)) = (891 , 0 , 0 , 0) . Derive the following solution for initial value problem y 1 ( t ) = 891exp( - λ 1 t ) y 2 ( t ) = 99exp( - λ 1 t ) - 99exp( - 10 λ 1 t ) y 3 ( t ) = - 1100exp( - λ 1 t ) + 100exp( - 10 λ 1 t ) + 1000exp( - 1 10 λ 1 t ) y 4 ( t ) = 110exp(
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Homework2Sol - Homework 2 AMATH 383, Autumn 2009 Due:...

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