Explanation Explanation The decay constant is calculated by the formula given below. λ = 0.693 t 1 / 2 Where t 1 / 2 is the half life of nuclide. λ is decay constant. The value of t 1 / 2 is 12.3 years . Substitute the value of half life in the above formula. λ = 0.693 12.3 year -1 = 0 . 0 5 6 3 4 y e a r s - 1 _ . The decay constant is 0 . 0 5 6 3 4 y e a r - 1 _ Explanation Decay time is calculated by the formula, t = 2.303 λ log ( n 0 n ) Where n 0 is the number of atoms initially present. n is the number of atoms remaining after decay. t is the decay time. Fraction of tritium remaining is 17 n 0 100 . Therefore, n = 17 n 0 100 Substitute the values of n , n 0 and decay constant in the above equation. t = 2

Student Solutions Manual for Zumdahl/Zumdahl/DeCoste?s Chemistry, 10th Edition

10th Edition

ISBN: 9781305957510

Author: ZUMDAHL, Steven S.; Zumdahl, Susan A.; DeCoste, Donald J.

Publisher: Cengage Learning

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 19, Problem 73AE

Interpretation Introduction

Interpretation: Use of tritium as a component of fluorescent watch dials and its half life is given. The year upto which the time at night can be read using the given kind of watch is to be calculated.

Concept introduction: A process through which an unstable nuclide looses its energy due to excess of protons or neutrons is known as radioactive decay. The cause of instability of a nuclide is its inefficiency in holding the nucleus together. Tritium was used as a component of fluorescent watch dials during world war 2

To determine: The year up to which the time at night can be read using the given kind of watch.

Expert Solution & Answer

Chapter 19 Solutions

Student Solutions Manual for Zumdahl/Zumdahl/DeCoste?s Chemistry, 10th Edition

Ch. 19 - Prob. 1RQ Ch. 19 - Prob. 2RQ Ch. 19 - Prob. 3RQ Ch. 19 - Prob. 4RQ Ch. 19 - Prob. 5RQ Ch. 19 - Prob. 6RQ Ch. 19 - Prob. 7RQ Ch. 19 - Prob. 8RQ Ch. 19 - Prob. 9RQ Ch. 19 - Prob. 10RQ

Ch. 19 - Prob. 1Q Ch. 19 - Prob. 3Q Ch. 19 - Prob. 4Q Ch. 19 - Prob. 5Q Ch. 19 - Prob. 6Q Ch. 19 - Prob. 7Q Ch. 19 - Prob. 8Q Ch. 19 - Prob. 9Q Ch. 19 - Prob. 10Q Ch. 19 - Prob. 11Q Ch. 19 - Prob. 12Q Ch. 19 - Prob. 13Q Ch. 19 - Prob. 14Q Ch. 19 - Prob. 15E Ch. 19 - Prob. 16E Ch. 19 - Prob. 17E Ch. 19 - Prob. 18E Ch. 19 - Prob. 19E Ch. 19 - Prob. 20E Ch. 19 - Prob. 21E Ch. 19 - Prob. 22E Ch. 19 - Prob. 23E Ch. 19 - Prob. 24E Ch. 19 - Prob. 27E Ch. 19 - Prob. 28E Ch. 19 - Prob. 29E Ch. 19 - Prob. 30E Ch. 19 - Prob. 32E Ch. 19 - Prob. 34E Ch. 19 - Prob. 35E Ch. 19 - Prob. 36E Ch. 19 - Prob. 37E Ch. 19 - Prob. 38E Ch. 19 - Prob. 39E Ch. 19 - Prob. 40E Ch. 19 - Prob. 41E Ch. 19 - Prob. 42E Ch. 19 - Prob. 43E Ch. 19 - Prob. 44E Ch. 19 - Prob. 45E Ch. 19 - Prob. 46E Ch. 19 - Prob. 47E Ch. 19 - Prob. 48E Ch. 19 - Prob. 49E Ch. 19 - Prob. 50E Ch. 19 - Prob. 52E Ch. 19 - Prob. 53E Ch. 19 - Prob. 54E Ch. 19 - Prob. 55E Ch. 19 - Prob. 56E Ch. 19 - Prob. 57E Ch. 19 - Prob. 58E Ch. 19 - Prob. 59E Ch. 19 - Prob. 60E Ch. 19 - Prob. 61E Ch. 19 - Prob. 62E Ch. 19 - Prob. 63E Ch. 19 - Prob. 64E Ch. 19 - Prob. 65AE Ch. 19 - Prob. 66AE Ch. 19 - Prob. 67AE Ch. 19 - Prob. 68AE Ch. 19 - Prob. 69AE Ch. 19 - Prob. 70AE Ch. 19 - Prob. 71AE Ch. 19 - Prob. 72AE Ch. 19 - Prob. 73AE Ch. 19 - Prob. 74AE Ch. 19 - Prob. 75AE Ch. 19 - Prob. 76AE Ch. 19 - Prob. 77AE Ch. 19 - Prob. 78AE Ch. 19 - Prob. 79AE Ch. 19 - Prob. 80AE Ch. 19 - Prob. 81CWP Ch. 19 - Prob. 82CWP Ch. 19 - Prob. 83CWP Ch. 19 - Prob. 84CWP Ch. 19 - Prob. 85CWP Ch. 19 - Prob. 86CWP Ch. 19 - Prob. 87CP Ch. 19 - Prob. 88CP Ch. 19 - Prob. 89CP Ch. 19 - Prob. 90CP Ch. 19 - Prob. 91CP Ch. 19 - Prob. 92CP Ch. 19 - Prob. 93CP Ch. 19 - Prob. 94CP Ch. 19 - Prob. 95IP Ch. 19 - Prob. 96IP

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
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Introductory Chemistry For Today
Chemistry
ISBN:9781285644561
Author:Seager
Publisher:Cengage
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning