** Subscribe** to view the full document.

**Unformatted text preview: **C This result can be reorganized as (6) Y H t L = C 1 ‰-k t Finally we use the initial condition to determine C 1 where Y H L = Y . The result is (7) Y H t L = Y ‰-k t The amounts of radioactive material at times at t 1 and t 2 are (8) Y H t 1 L = Y ‰-k t 1 , Y H t 2 L = Y ‰-k t 2 Then in order that the Y H t 2 L is half that at Y H t 1 L we have (9) Y ‰-k t 2 = 1 ÅÅÅÅ 2 Y ‰-k t 1 Simplifying this expression gives (10) ‰-k H t 2-t 1 L = 1 ÅÅÅÅ 2 Solving for t 2-t 1 gives (11) T = H t 2-t 1 L = Ln H 2 L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ k If T = 5568 years, then k is (12) k = Ln H 2 L ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 5568 = 0.000124488 yrs-1 2 Problem_1.8.nb...

View
Full Document

- Summer '13
- LeThuy
- Radioactive Decay, Half-Life, Y0 ‰-k t1, Ln H2L