Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
Book Icon
Chapter 21, Problem 21.82QA
Interpretation Introduction

To find:

a)  The percent difference in the carbon-14 decay rate in biological samples from 1360 BCE and 1628 BCE.

b) The ratio of carbon-14: carbon-12 in the blackened grains compared with that of the grain harvested last year.

Expert Solution & Answer
Check Mark

Answer to Problem 21.82QA

Solution:

a)  The percent difference in the carbon-14 decay rate in biological samples from 1360 BCE and 1628 BCE = 3%

b)  The ratio of carbon-14: carbon-12 in the blackened grains from the Jericho site (1315 ±13  years) compared with that of the grain harvested last year is 1: 0.27

Explanation of Solution

1) Concept:

From the half-life of nuclear decay, and the ratio of isotopes, we can calculate the isotopic ratios backwards in time. This is a powerful tool for radioactive dating techniques. This technique is used extensively for geological dating of various fossils, and rocks.

Ratio of C-14 to C-12 is a powerful tool used for radioactive dating..

2) Formula:

i) t12=0.693k

ii) lnXtX0=-kt

where, Xt is the final concentration, X0 is initial concentration, k is rate constant,  t is time, and t12 is half-life

3) Given:

i) t1=1360 years,  t2=1628 years

ii) t=1315±13 years      

4) Calculations:

The half-life of C 14 is taken from the Table 21.2 as 5730 years.

So, the rate constant is calculated as follows:

t12=0.693k

5730 yr=0.693k

k=0.6935730 yr=1.209×10-4yr-1

Part a)

Calculation using t1=1360 years:

 lnXtX0=-kt

XtX0=e-kt=e-1.209×10-4yr-1×1360 yr=0.84

XtX0=85%

Calculation using t2=1628 years:

 lnXtX0=-kt

XtX0=e-kt=e-1.209×10-4yr-1×1628 yr=0.82

XtX0=82%

Therefore, the percent difference is  85%-82%=3%

Thus the percent difference in the carbon-14 decay rate in biological samples from 1360 BCE and 1628 BCE = 3%

Part b)

t=1315±13 years

So, t1=1315-13=1302 years  and  t2=1315+13=1328 years

Calculation using t1=1302 years:

 lnXtX0=-kt

XtX0=e-kt=e-1.209×10-4yr-1×1302 yr=0.8543

XtX0=85.43%

Calculation using t2=1328 years:

 lnXtX0=-kt

XtX0=e-kt=e-1.209×10-4yr-1×1328 yr=0.8516

XtX0=85.16%

Therefore, the percent difference is 85.43%-85.16%=0.27%

For grain harvested last year, take t = 1 year.

k = 1.209×10-4yr-1

XtX0=e-kt= 0.999

Thus the ratio of carbon-14: carbon-12 in the blackened grains from the Jericho site (1315 ±13  years) compared with that of the grain harvested last year is 1: 0.27

Conclusion:

The ratio of carbon-14: carbon-12 is calculated using the first order radioactive decay equation.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Part A Give the IUPAC name of the amine shown. CH3-CH2 Spell out the full name of the compound. N CH2 CH2-CH3 CH2 CH2-CH3 Submit Hints My Answers Give Up Review Part
Show work. don't give Ai generated solution
None

Chapter 21 Solutions

Chemistry: An Atoms-Focused Approach

Ch. 21 - Prob. 21.12VPCh. 21 - Prob. 21.13QACh. 21 - Prob. 21.14QACh. 21 - Prob. 21.15QACh. 21 - Prob. 21.16QACh. 21 - Prob. 21.17QACh. 21 - Prob. 21.18QACh. 21 - Prob. 21.19QACh. 21 - Prob. 21.20QACh. 21 - Prob. 21.22QACh. 21 - Prob. 21.23QACh. 21 - Prob. 21.24QACh. 21 - Prob. 21.25QACh. 21 - Prob. 21.26QACh. 21 - Prob. 21.27QACh. 21 - Prob. 21.28QACh. 21 - Prob. 21.29QACh. 21 - Prob. 21.30QACh. 21 - Prob. 21.31QACh. 21 - Prob. 21.32QACh. 21 - Prob. 21.33QACh. 21 - Prob. 21.34QACh. 21 - Prob. 21.35QACh. 21 - Prob. 21.36QACh. 21 - Prob. 21.37QACh. 21 - Prob. 21.38QACh. 21 - Prob. 21.39QACh. 21 - Prob. 21.40QACh. 21 - Prob. 21.41QACh. 21 - Prob. 21.42QACh. 21 - Prob. 21.43QACh. 21 - Prob. 21.44QACh. 21 - Prob. 21.45QACh. 21 - Prob. 21.46QACh. 21 - Prob. 21.47QACh. 21 - Prob. 21.48QACh. 21 - Prob. 21.49QACh. 21 - Prob. 21.50QACh. 21 - Prob. 21.51QACh. 21 - Prob. 21.52QACh. 21 - Prob. 21.53QACh. 21 - Prob. 21.54QACh. 21 - Prob. 21.55QACh. 21 - Prob. 21.56QACh. 21 - Prob. 21.57QACh. 21 - Prob. 21.58QACh. 21 - Prob. 21.59QACh. 21 - Prob. 21.60QACh. 21 - Prob. 21.61QACh. 21 - Prob. 21.62QACh. 21 - Prob. 21.63QACh. 21 - Prob. 21.64QACh. 21 - Prob. 21.65QACh. 21 - Prob. 21.66QACh. 21 - Prob. 21.67QACh. 21 - Prob. 21.68QACh. 21 - Prob. 21.69QACh. 21 - Prob. 21.70QACh. 21 - Prob. 21.71QACh. 21 - Prob. 21.72QACh. 21 - Prob. 21.73QACh. 21 - Prob. 21.74QACh. 21 - Prob. 21.75QACh. 21 - Prob. 21.76QACh. 21 - Prob. 21.77QACh. 21 - Prob. 21.78QACh. 21 - Prob. 21.79QACh. 21 - Prob. 21.80QACh. 21 - Prob. 21.81QACh. 21 - Prob. 21.82QACh. 21 - Prob. 21.83QACh. 21 - Prob. 21.84QACh. 21 - Prob. 21.85QACh. 21 - Prob. 21.86QACh. 21 - Prob. 21.87QACh. 21 - Prob. 21.88QACh. 21 - Prob. 21.89QACh. 21 - Prob. 21.90QACh. 21 - Prob. 21.91QACh. 21 - Prob. 21.92QACh. 21 - Prob. 21.93QACh. 21 - Prob. 21.94QACh. 21 - Prob. 21.95QACh. 21 - Prob. 21.96QACh. 21 - Prob. 21.97QACh. 21 - Prob. 21.98QACh. 21 - Prob. 21.99QA
Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY