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Chapter_23.3_23.6_student version

# Chapter_23.3_23.6_student version - Radioactive Decay We...

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Radioactive Decay We saw that: N = N 0 exp(- λ t) 23.3 Substituting: N/N 0 = exp(-(ln 2)t/t 1/2 ) and: λ = (ln 2)/t 1/2 Rearranging: N/N 0 = exp(-(t/t 1/2 )ln 2) Rearranging: N/N 0 = exp(ln 2 ) -(t/t 1/2 ) exp(ln x) = x, so: N/N 0 = 2 -(t/t 1/2 )

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Radioactive Decay 23.3 N/N 0 = 2 -(t/t 1/2 ) Example: How much of a radioisotope remains after: (a) 1 half life? (b) 2 half lives? (c) 5 half lives? (a) after one half life, t = t 1/2 N/N 0 = 2 -(t/t 1/2 ) (b) after two half lives, t = 2 t 1/2 N/N 0 = 2 -(t/t 1/2 ) (c) after five half lives, t = 5 t 1/2 Note that abundance goes to zero at infinity !
Radioactive Decay 23.3 N/N 0 = 2 -(t/t 1/2 ) Example: How much 14 C is left after 15,000 years if the half life is 5730 years? N/N 0 N/N 0 N/N 0 Therefore 16.3 % of the sample of 14 C remains after 15,000 years of decay.

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Dating Based on Radioactive Decay 23.3 bring t’s down: ln(N/N 0 ) = - (t/t 1/2 )ln 2 take ln: ln(N/N 0 ) = ln 2 - (t/t 1/2 ) We saw that: N/N 0 = 2 - (t/t 1/2 ) But what if we know the amount remaining and want to calculate the age of the specimen? We need to rearrange the equation to calculate t.
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Chapter_23.3_23.6_student version - Radioactive Decay We...

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