# A living plant contains approximately the same fraction of carbon-14 as in atmospheric carbon dioxide. Assuming that the observed rate of decay of carbon-14 from a living plant is 13.6 counts per minute per gram of carbon, how many counts per minute per gram of carbon will be measured from a 15,000-year-old sample? Will radiocarbon dating work well for small samples of 10 mg or less? (For 14 C, t 1/2 = 5730 years.)

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

Chapter
Section

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 18, Problem 37E
Textbook Problem
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## A living plant contains approximately the same fraction of carbon-14 as in atmospheric carbon dioxide. Assuming that the observed rate of decay of carbon-14 from a living plant is 13.6 counts per minute per gram of carbon, how many counts per minute per gram of carbon will be measured from a 15,000-year-old sample? Will radiocarbon dating work well for small samples of 10 mg or less? (For 14C, t1/2 = 5730 years.)

Interpretation Introduction

Interpretation: Rate of decay of carbon 14 and its half life is given. The counts per minute per gram of carbon measured from a 15000 year old sample are to be determined. Effectiveness of radiocarbon dating for a small sample of 10mg or less is to be justified.

Concept introduction: Radiocarbon dating is based on the radioactivity of the nuclide 614C which decays by beta particle production.

A living plant consumes carbon dioxide in the photosynthesis process and incorporates the carbon including 614C into its molecule.

To determine: The counts per minute per gram of carbon measured from a 15000 year old sample.

### Explanation of Solution

Explanation

The decay constant can be calculated by the formula given below.

λ=0.693t1/2

Where

• t1/2 is the half life of nuclide.
• λ is the decay constant.

Substitute the value of t1/2 in the above equation.

λ=0.6935730yrs1λ=1.2094×10-4yrs-1_

Explanation

The counts per minute per gram are calculated by the formula,

t=2.303λlog(OriginalactivityFinalactivity)

Substitute the value of λ , the original activity and the final activity in the above expression

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