Explanation If we consider after starting the radioactive disintegration of a certain nucleus after the time gap ‘t’ the number of atom disintegrate is ‘N’. Then for small gap of time ‘dt’ the disintegration of the atom will be ‘dN’. Then as per the law of radioactive disintegration the rate of decay ( − d N d t ) of the atom will be equivalent to the number of atom present at the moment. so, − d N d t ∞ N or, − d N d t = λ N (Where λ is a constant called disintegration constant or decay constant). The value of the decay constant is fixed for a radioactive material. The half life time of a radioactive material is the time required to disintegrate half of the material. It is expressed by: t 1 / 2 = 0.693 λ Given Info: The half-life of 238 U is 4.5 × 10 9 yr. The sample of rock of mass 1.6g produces 29 dis/s. Formula used: − d N d t = λ N , t 1 / 2 = 0.693 λ , 1curie (C) = 3.7 × 10 10 dis/s or dps Calculation: We know 1curie(C) = 3.7 × 10 10 dis/s or dps,so, 29 dps = 29 3.7 × 10 = 7.837 × 10 − 10 C So, the rate of dissociation − d N d t = 7.837 × 10 − 10 …………Equation 1 The dissociation constant λ = 0.693 t 1 2 so, from the given data the dissociation constant is λ = 0...

1st Edition

ISBN: 9780321834683

Author: Nivaldo J. Tro

Publisher: PEARSON

See similar textbooks

Concept explainers

Writing and Balancing Nuclear Equations

A nuclear equation is a symbolic representation of a nuclear reaction using certain symbols. It is studied in nuclear chemistry and nuclear physics.

Question

Chapter 21, Problem 91E

Interpretation Introduction

To determine: The percent of mass of ²³⁸U in the rock.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 21, Problem 90E

Chapter 21, Problem 92E

Chapter 21 Solutions

Chemistry: Structure and Properties

Ch. 21 - Prob. 1SAQ Ch. 21 - Prob. 2SAQ Ch. 21 - Prob. 3SAQ Ch. 21 - Prob. 4SAQ Ch. 21 - Prob. 5SAQ Ch. 21 - Prob. 6SAQ Ch. 21 - Prob. 7SAQ Ch. 21 - Prob. 8SAQ Ch. 21 - Prob. 9SAQ Ch. 21 - Prob. 10SAQ

Ch. 21 - Prob. 1E Ch. 21 - Prob. 2E Ch. 21 - Prob. 3E Ch. 21 - Prob. 4E Ch. 21 - Prob. 5E Ch. 21 - Prob. 6E Ch. 21 - Prob. 7E Ch. 21 - Prob. 8E Ch. 21 - Prob. 9E Ch. 21 - Prob. 10E Ch. 21 - Prob. 11E Ch. 21 - Prob. 12E Ch. 21 - Prob. 13E Ch. 21 - Prob. 14E Ch. 21 - Prob. 15E Ch. 21 - Prob. 16E Ch. 21 - Prob. 17E Ch. 21 - Prob. 18E Ch. 21 - Prob. 19E Ch. 21 - Prob. 20E Ch. 21 - Prob. 21E Ch. 21 - Prob. 22E Ch. 21 - Prob. 23E Ch. 21 - Prob. 24E Ch. 21 - Prob. 25E Ch. 21 - Prob. 26E Ch. 21 - Prob. 27E Ch. 21 - Prob. 28E Ch. 21 - Prob. 29E Ch. 21 - Prob. 30E Ch. 21 - Prob. 31E Ch. 21 - Prob. 32E Ch. 21 - Prob. 33E Ch. 21 - Prob. 34E Ch. 21 - Prob. 35E Ch. 21 - Prob. 36E Ch. 21 - Prob. 37E Ch. 21 - Prob. 38E Ch. 21 - Prob. 39E Ch. 21 - Prob. 40E Ch. 21 - Prob. 41E Ch. 21 - Prob. 42E Ch. 21 - Prob. 43E Ch. 21 - Prob. 44E Ch. 21 - Prob. 45E Ch. 21 - Prob. 46E Ch. 21 - Prob. 47E Ch. 21 - Prob. 48E Ch. 21 - Prob. 49E Ch. 21 - Prob. 50E Ch. 21 - Prob. 51E Ch. 21 - Prob. 52E Ch. 21 - Prob. 53E Ch. 21 - Prob. 54E Ch. 21 - Prob. 55E Ch. 21 - Prob. 56E Ch. 21 - Prob. 57E Ch. 21 - Prob. 58E Ch. 21 - Prob. 59E Ch. 21 - Prob. 60E Ch. 21 - Prob. 61E Ch. 21 - Prob. 62E Ch. 21 - Prob. 63E Ch. 21 - Prob. 64E Ch. 21 - Prob. 65E Ch. 21 - Prob. 66E Ch. 21 - Prob. 67E Ch. 21 - Prob. 68E Ch. 21 - Prob. 69E Ch. 21 - Prob. 70E Ch. 21 - Prob. 71E Ch. 21 - Prob. 72E Ch. 21 - Prob. 73E Ch. 21 - Prob. 74E Ch. 21 - Prob. 75E Ch. 21 - Prob. 76E Ch. 21 - Prob. 77E Ch. 21 - Prob. 78E Ch. 21 - Prob. 79E Ch. 21 - Prob. 80E Ch. 21 - Prob. 81E Ch. 21 - Prob. 82E Ch. 21 - Prob. 83E Ch. 21 - Prob. 84E Ch. 21 - Prob. 85E Ch. 21 - Prob. 86E Ch. 21 - Prob. 87E Ch. 21 - Prob. 88E Ch. 21 - Prob. 89E Ch. 21 - Prob. 90E Ch. 21 - Prob. 91E Ch. 21 - Prob. 92E Ch. 21 - Prob. 93E Ch. 21 - Prob. 94E Ch. 21 - Prob. 95E Ch. 21 - Prob. 96E Ch. 21 - Prob. 97E Ch. 21 - Prob. 98E Ch. 21 - Prob. 99E Ch. 21 - Prob. 100E Ch. 21 - Prob. 101E Ch. 21 - Prob. 102E Ch. 21 - Prob. 103E Ch. 21 - Prob. 104E Ch. 21 - Prob. 105E Ch. 21 - Prob. 106E Ch. 21 - Prob. 107E Ch. 21 - Prob. 108E Ch. 21 - Prob. 109E

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions

A sample of rock was found to contain 8.23 mg of rubidium-87 and 0.47 mg of strontium-87.. (a) Calculate the age of the rock if the half-life of the decay of rubidium by emission is 4.71010 y. (b) If some S3887r was initially present in the rock, would the rock be younger, older, or the same age as the age calculated in (a)? Explain your answer.

Recommended textbooks for you

Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Introductory Chemistry For Today
Chemistry
ISBN:9781285644561
Author:Seager
Publisher:Cengage
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning

Chemistry: Principles and Reactions

Chemistry

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:Cengage Learning

Introductory Chemistry For Today

Chemistry

ISBN:9781285644561

Author:Seager

Publisher:Cengage

General Chemistry - Standalone book (MindTap Cour...

Chemistry

ISBN:9781305580343

Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell

Publisher:Cengage Learning

Chemistry & Chemical Reactivity

Chemistry

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:Cengage Learning

Chemistry & Chemical Reactivity

Chemistry

ISBN:9781133949640

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:Cengage Learning

Chemistry

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Cengage Learning

SEE MORE TEXTBOOKS

The percent of mass of 238 U in the rock.

Solution Summary: The author calculates the percent of mass of 238 U in the rock sample.

Concept explainers

To determine: The percent of mass of ²³⁸U in the rock.

Want to see the full answer?

Chapter 21 Solutions

The percent of mass of 238 U in the rock.

Solution Summary: The author calculates the percent of mass of 238 U in the rock sample.

Concept explainers

To determine: The percent of mass of 238U in the rock.

Want to see the full answer?

Chapter 21 Solutions

To determine: The percent of mass of ²³⁸U in the rock.