p43_076

p43_076 - state computed in part(b Thus the ratio is(0...

This preview shows page 1. Sign up to view the full content.

76. (a) From the decay series, we know that N 210 , the amount of 210 Pb nuclei, changes because of two decays: the decay from 226 Ra into 210 Pb at the rate R 226 = λ 226 N 226 , and the decay from 210 Pb into 206 Pb at the rate R 210 = λ 210 N 210 . The ±rst of these decays causes N 210 to increase while the second one causes it to decrease. Thus, dN 210 dt = R 226 R 210 = λ 226 N 226 λ 210 N 210 . (b) We set dN 210 /dt = R 226 R 210 = 0 to obtain R 226 /R 210 =1. (c) From R 226 = λ 226 N 226 = R 210 = λ 210 N 210 ,weobta in N 226 N 210 = λ 210 λ 226 = T 1 / 2 226 T 1 / 2 210 = 1 . 60 × 10 3 y 22 . 6y =70 . 8 . (d) Since only 1 . 00% of the 226 Ra remains, the ratio R 226 /R 210 is 0 .
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: state computed in part (b). Thus the ratio is (0 . 0100)(1) = 0 . 0100. (e) This is similar to part (d) above. Since only 1 . 00% of the 226 Ra remains, the ratio N 226 /N 210 is 1 . 00% of that of the equilibrium state computed in part (c), or (0 . 0100)(70 . 8) = 0 . 708. (f) Since the actual value of N 226 /N 210 is 0.09, which much closer to 0.0100 than to 1, the sample of the lead pigment cannot be 300 years old. So Emmaus is not a Vermeer ....
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online