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

9th Edition

ISBN: 9781133611097

Author: Steven S. Zumdahl

Publisher: Cengage Learning

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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 65AE

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

Chemistry

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. 2Q 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. 11E Ch. 19 - Prob. 12E Ch. 19 - Prob. 13E Ch. 19 - Prob. 14E 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. 23E Ch. 19 - Prob. 24E Ch. 19 - Prob. 26E Ch. 19 - Prob. 27E Ch. 19 - Prob. 28E Ch. 19 - Prob. 29E Ch. 19 - Prob. 30E Ch. 19 - Prob. 31E Ch. 19 - Prob. 32E Ch. 19 - Prob. 33E 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. 46E Ch. 19 - Prob. 47E Ch. 19 - Prob. 48E Ch. 19 - Prob. 49E Ch. 19 - Prob. 50E Ch. 19 - Prob. 51E 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. 59AE Ch. 19 - Prob. 60AE Ch. 19 - Prob. 61AE Ch. 19 - Prob. 62AE Ch. 19 - Prob. 63AE Ch. 19 - Prob. 64AE 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. 73CWP Ch. 19 - Prob. 74CWP Ch. 19 - Prob. 75CWP Ch. 19 - Prob. 76CWP Ch. 19 - Prob. 77CWP Ch. 19 - Prob. 78CWP Ch. 19 - Prob. 79CP Ch. 19 - Prob. 80CP Ch. 19 - Prob. 81CP Ch. 19 - Prob. 82CP Ch. 19 - Prob. 83CP Ch. 19 - Prob. 84CP Ch. 19 - Prob. 85CP Ch. 19 - Prob. 86CP Ch. 19 - Prob. 87IP Ch. 19 - Prob. 88IP

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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.

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