Homework2Sol

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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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(
This is the end of the preview. Sign up to access the rest of the document.

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online