p43_031 - N = N 2 at the end of two half-lives N = N 4 and...

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31. (a) The half-life T 1 / 2 and the disintegration constant are related by T 1 / 2 =( l n2 ) ,s o T 1 / 2 = (ln 2) / (0 . 0108 h 1 )=64 . 2h. (b) At time t , the number of undecayed nuclei remaining is given by N = N 0 e λt = N 0 e (ln 2) t/T 1 / 2 . We substitute t =3 T 1 / 2 to obtain N N 0 = e 3ln2 =0
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Unformatted text preview: N = N / 2, at the end of two half-lives, N = N / 4, and at the end of three half-lives, N = N / 8 = . 125 N . (c) We use N = N e − λt . 10 . d is 240 h, so λt = (0 . 0108 h − 1 )(240 h) = 2 . 592 and N N = e − 2 . 592 = 0 . 0749 ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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