Iron-59 has a half-life of 45.1 days. How old is an iron nail if the Fe-59 content is 25% that of a new sample of iron? Show all calculations leading to a solution.

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Iron-59 has a half-life of 45.1 days. How old is an iron nail if the Fe-59 content is 25% that of a new sample of iron? Show all calculations leading to a solution.

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Answer 1 · 2017-08-22T00:22:11

90 days

Explanation:

All radio decay follows 1st order kinetics and therefore is supported by the integrated rate law of a 1st order decay trend. The classic form of the 1st order decay equation is $A = A_{o} e^{- kt}$ $A = A_oe^-"kt"$ where $A$ $A$ = final activity (or mass), $A_{o}$ $A_o$ = initial activity (or mass), $k$ $k$ = the rate constant & $t$ $t$ = time of decay. The rate constant $(k)$ $(k)$ as a function of half-life can be determined from $k \cdot t_{\frac{1}{2}} = 0.693$ $k*t_(1/2) = 0.693$ .

Given $t_{\frac{1}{2}} = 45.1 d a y s$ $t_(1/2) = 45.1 days$ => $k = (\frac{0.693}{45.1}) d a y s^{- 1}$ $k = (0.693/45.1)days^-1$ = $0.0154 d a y s^{- 1}$ $0.0154 days^-1$

From the 1st order decay equation ...
$A = A_{o} e^{- kt}$ $A = A_oe^-"kt"$ => $ln (\frac{A}{A_{o}}) = - k \cdot t$ $ln(A/A_o) = - k*t$
=> $t = ⎛ ⎜ ⎝ \frac{ln (\frac{A}{A_{o}})}{- k} ⎞ ⎟ ⎠$ $t = (ln(A/A_o)/-k)$ = $⎛ ⎜ ⎝ \frac{ln (\frac{25}{100})}{- 0.0154} ⎞ ⎟ ⎠ d a y s$ $(ln(25/100)/-0.0154)days$ = $90 d a y s$ $90days$

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Iron-59 has a half-life of 45.1 days. How old is an iron nail if the Fe-59 content is 25% that of a new sample of iron? Show all calculations leading to a solution.

1 Answer

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