Chemistry: Atoms First
Chemistry: Atoms First
2nd Edition
ISBN: 9780073511184
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
Question
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Chapter 20.3, Problem 20.3.2SR
Interpretation Introduction

Interpretation: From the age of the given wooden artifact, 14C activity should be determined.

Concept Introduction:

  • Unstable nuclei emit radiation spontaneously to become stable nuclei by losing energy. This process of emission of radiation by unstable nuclei is known as radioactive decay.
  • These emitted radiations may be alpha radiations(α), beta radiations(β) or gamma radiations(γ)
  • These unstable nuclei are the nuclei with more than 83 protons and which do not lie within the belt of stability.
  • Radioactive decay are in the first order kinetics. Rate of radioactive decay at a time t is,

Rate of decay at a time 't' = k Nk-Firstorderrateconstant and its unit is t1N-numberofradioactivenucleipresentattime't

Suppose the number of radioactive nuclei at time zero is N0 and at a time t is Nt,

lnNtN0=-kt

  • Half-life of radioactive decay is the time required for a radioactive sample to decay to one half of the atomic nucleus.

Half-life of the radiation,  t1/2=0.693k

Half-life and rate constant for radioactive isotopes vary greatly from nucleus to nucleus.

Expert Solution & Answer
Check Mark

Answer to Problem 20.3.2SR

 14C Activity of wooden artifact is  0.935disintegrationper second

Explanation of Solution

Determine the value of k

Half-life of radioactive decay is the time required for a radioactive sample to decay to one half of the atomic nucleus. From the half-life of the reaction, the value of k can be easily determined.

t1/2 =0.693kk =0.693t1/2

Here the half lie of carbon-14 is 5715 years. So,

k =0.6935715 years = 1.21×10-4yr-1

Determine the 14C activity of old wooden artifact in a 23000 years old

Suppose the number of radioactive nuclei at time zero is N0 and at a time t is Nt,

lnNtN0=-kt

Here 14C activity of the fresh cut wood is given. Since the radioactive sample is proportional to the number of radioactive nuclei, the above equation can be written with activity in place of concentration.

ln14Cactivityinoldwoodenobject14Cactivityinfresscutwood=-kt

From the given data's14C activityin fresh cut wood = 15.2disintegrationper second14C activityinthewoodenartifact = ?k = 1.21×10-4yr-1Ageofthewooden artifact,t =23,000yrs

By adding these two values,

ln14Cactivityinoldwoodenobject15.2disintegrationper second =-1.21×10-4yr-1×23000yr =- 2.78314Cactivityinwoodenartifact15.2dps =e-2.78314Cactivityinthewoodenartifact =e-2.783×15.2 dps   =0.940disintegrationper second

Conclusion

From the given details of the wooden object, 14C activity is determined by using the equation of rate of decay.

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Chapter 20 Solutions

Chemistry: Atoms First

Ch. 20.2 - Prob. 20.2.1SRCh. 20.2 - Prob. 20.2.2SRCh. 20.2 - Prob. 20.2.3SRCh. 20.2 - Prob. 20.2.4SRCh. 20.3 - Prob. 20.3WECh. 20.3 - Prob. 3PPACh. 20.3 - Prob. 3PPBCh. 20.3 - Prob. 3PPCCh. 20.3 - Prob. 20.4WECh. 20.3 - Prob. 4PPACh. 20.3 - Prob. 4PPBCh. 20.3 - Prob. 4PPCCh. 20.3 - Prob. 20.3.1SRCh. 20.3 - Prob. 20.3.2SRCh. 20.3 - Prob. 20.3.3SRCh. 20.4 - Prob. 20.5WECh. 20.4 - Prob. 5PPACh. 20.4 - Prob. 5PPBCh. 20.4 - Prob. 5PPCCh. 20.4 - Prob. 20.4.1SRCh. 20.4 - Prob. 20.4.2SRCh. 20 - Prob. 20.1QPCh. 20 - Prob. 20.2QPCh. 20 - Prob. 20.3QPCh. 20 - Prob. 20.4QPCh. 20 - Prob. 20.5QPCh. 20 - Prob. 20.6QPCh. 20 - Prob. 20.7QPCh. 20 - Prob. 20.8QPCh. 20 - Prob. 20.9QPCh. 20 - Prob. 20.10QPCh. 20 - Prob. 20.11QPCh. 20 - Prob. 20.12QPCh. 20 - Prob. 20.13QPCh. 20 - Prob. 20.14QPCh. 20 - Prob. 20.15QPCh. 20 - Prob. 20.16QPCh. 20 - Prob. 20.17QPCh. 20 - Prob. 20.18QPCh. 20 - Prob. 20.19QPCh. 20 - Prob. 20.20QPCh. 20 - Prob. 20.21QPCh. 20 - Prob. 20.22QPCh. 20 - Prob. 20.23QPCh. 20 - Prob. 20.24QPCh. 20 - Prob. 20.25QPCh. 20 - Prob. 20.26QPCh. 20 - Prob. 20.27QPCh. 20 - Prob. 20.28QPCh. 20 - Prob. 20.29QPCh. 20 - Prob. 20.30QPCh. 20 - Prob. 20.31QPCh. 20 - Prob. 20.32QPCh. 20 - Prob. 20.33QPCh. 20 - Prob. 20.34QPCh. 20 - Prob. 20.35QPCh. 20 - Prob. 20.36QPCh. 20 - Prob. 20.37QPCh. 20 - Prob. 20.38QPCh. 20 - Prob. 20.39QPCh. 20 - Prob. 20.1VCCh. 20 - Prob. 20.2VCCh. 20 - Prob. 20.3VCCh. 20 - Prob. 20.4VCCh. 20 - Prob. 20.40QPCh. 20 - Prob. 20.41QPCh. 20 - Prob. 20.42QPCh. 20 - Prob. 20.43QPCh. 20 - Prob. 20.44QPCh. 20 - Prob. 20.45QPCh. 20 - Prob. 20.46QPCh. 20 - Prob. 20.47QPCh. 20 - Prob. 20.48QPCh. 20 - Prob. 20.49QPCh. 20 - Prob. 20.50QPCh. 20 - Prob. 20.51QPCh. 20 - Prob. 20.52QPCh. 20 - Prob. 20.53QPCh. 20 - Prob. 20.54QPCh. 20 - Prob. 20.55QPCh. 20 - Prob. 20.56QPCh. 20 - Prob. 20.57QPCh. 20 - Prob. 20.58QPCh. 20 - Prob. 20.59QPCh. 20 - Prob. 20.60QPCh. 20 - Prob. 20.61QPCh. 20 - Prob. 20.62QPCh. 20 - Prob. 20.63QPCh. 20 - Prob. 20.64QPCh. 20 - Prob. 20.65QPCh. 20 - Prob. 20.66QPCh. 20 - Prob. 20.67QPCh. 20 - Prob. 20.68QPCh. 20 - Prob. 20.69QPCh. 20 - Prob. 20.70QPCh. 20 - Prob. 20.71QPCh. 20 - Prob. 20.72QPCh. 20 - Prob. 20.73QPCh. 20 - Prob. 20.74QPCh. 20 - Prob. 20.75QPCh. 20 - Prob. 20.76QPCh. 20 - Prob. 20.77QPCh. 20 - Prob. 20.78QPCh. 20 - Prob. 20.79QPCh. 20 - Prob. 20.80QPCh. 20 - Prob. 20.81QPCh. 20 - Prob. 20.82QPCh. 20 - Prob. 20.83QPCh. 20 - Prob. 20.84QPCh. 20 - Prob. 20.85QPCh. 20 - Prob. 20.86QPCh. 20 - Prob. 20.87QPCh. 20 - Prob. 20.88QPCh. 20 - Prob. 20.89QPCh. 20 - Prob. 20.90QPCh. 20 - Prob. 20.91QPCh. 20 - Prob. 20.92QPCh. 20 - Prob. 20.93QPCh. 20 - Prob. 20.94QPCh. 20 - Prob. 20.95QPCh. 20 - Prob. 20.96QPCh. 20 - Prob. 20.97QPCh. 20 - Prob. 20.98QPCh. 20 - Prob. 20.99QPCh. 20 - Prob. 20.100QPCh. 20 - Prob. 20.101QP
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