Carbon from a glacier mummy discovered in the Tyrolean Alps has an activity of 5.47 decays of- 14C per minute per gram of carbon. If carbon from a living organism in this region has an activity of 13.5 decays/min/g C, estimate the age of the mummy. (The half-life of ¹4C is 5730 years.) age = years old

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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**Question 7**

Carbon from a glacier mummy discovered in the Tyrolean Alps has an activity of 5.47 decays of \(^{14}C\) per minute per gram of carbon. If carbon from a living organism in this region has an activity of 13.5 decays/min/g C, estimate the age of the mummy. (The half-life of \(^{14}C\) is 5730 years.)

age = [ ] years old

**Explanation:**

To estimate the age of the mummy, use the formula for radioactive decay:

\[ N(t) = N_0 \times (1/2)^{t/T_{1/2}} \]

Where:
- \( N(t) \) is the current activity (5.47 decays/min/g),
- \( N_0 \) is the initial activity (13.5 decays/min/g),
- \( t \) is the time (age of the mummy),
- \( T_{1/2} \) is the half-life of \(^{14}C\) (5730 years).

Rearrange to solve for \( t \):

\[ t = \frac{\log(N(t)/N_0)}{\log(1/2)} \times T_{1/2} \]
Transcribed Image Text:**Question 7** Carbon from a glacier mummy discovered in the Tyrolean Alps has an activity of 5.47 decays of \(^{14}C\) per minute per gram of carbon. If carbon from a living organism in this region has an activity of 13.5 decays/min/g C, estimate the age of the mummy. (The half-life of \(^{14}C\) is 5730 years.) age = [ ] years old **Explanation:** To estimate the age of the mummy, use the formula for radioactive decay: \[ N(t) = N_0 \times (1/2)^{t/T_{1/2}} \] Where: - \( N(t) \) is the current activity (5.47 decays/min/g), - \( N_0 \) is the initial activity (13.5 decays/min/g), - \( t \) is the time (age of the mummy), - \( T_{1/2} \) is the half-life of \(^{14}C\) (5730 years). Rearrange to solve for \( t \): \[ t = \frac{\log(N(t)/N_0)}{\log(1/2)} \times T_{1/2} \]
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