Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Chapter 18, Problem 79CP
Interpretation Introduction

Interpretation: Half life of 238U and 235U and their relative abundances is given. Their relative abundances when earth was formed 4.5 billion years ago are 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. Decay constant is the quantity that expresses the rate of decrease of number of atoms of a radioactive element per second. Half life of radioactive sample is defined as the time required for the number of nuclides to reach half of the original value.

To determine: The relative abundances of 238U and 235U when earth was formed.

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Answer to Problem 79CP

Answer

The relative abundance of 238U when earth was formed is 76%_ .

The relative abundance of 235U when earth was formed is 24%_ .

Explanation of Solution

Explanation

The decay constant is calculated by the formula given below.

λ=0.693t1/2

Where

  • t1/2 is the half life of nuclide.
  • λ is decay constant.

The value of t1/2 for 238U is 4.5×109years .

Substitute the value of half life in the above formula.

λ=0.6934.5×109years-1

The decay constant is 0.6934.5×109years-1 .

The decay constant is calculated by the formula given below.

λ=0.693t1/2

Where

  • t1/2 is the half life of nuclide.
  • λ is decay constant.

The value of t1/2 for 235U is 7.1×108years .

Substitute the value of half life in the above formula.

λ=0.6937.1×108years-1 .

The decay constant is 0.6937.1×108years-1 .

The relative abundance of 238U is calculated by the formula,

ln(nn0)=λt

Where

  • n0 is the amount of 238U when earth was formed.
  • n is the amount of 238U after 4.5×109 years.
  • t is the decay time.

The value of t is 4.5×109 years.

Substitute the values of t and decay constant in the above equation.

ln(nn0)=λtln(nn0)=0.693×4.5×1094.5×109ln(nn0)=0.693(nn0)=0.50

The value of (nn0) for 238U is 0.50 .

The relative abundance of 235U is calculated by the formula,

ln(nn0)=λt

Where

  • n0 is the amount of 235U when earth was formed.
  • n is the amount of 235U after 4.5×109 years.
  • t is the decay time.

The value of t is 7.1×108 years.

Substitute the values of t and decay constant in the above equation.

ln(nn0)=λtln(nn0)=0.693×4.5×1097.1×108(nn0)=e4.39(nn0)=0.012

The value of (nn0) for 235U is 0.012 .

The natural composition of 238U is 99.28%=99281000

Therefore, 9928 nuclei of 238U are present in 10000 uranium nuclei.

Therefore, the value of n for 238U is 9928 .

Substitute the value of n in the equation,

(nn0)=0.50n0=n0.50n0=99280.50=1.9×104238Unuclei

The amount of 238U when earth was formed is 1.9×104238Unuclei

The natural composition of 235U is 0.72%=721000

Therefore, 72 nuclei of 235U are present in 10000 uranium nuclei.

Therefore, the value of n for 235U is 72 .

Substitute the value of n in the equation,

(nn0)=0.012n0=n0.012n0=720.012=6×103235Unuclei

The amount of 235U when earth was formed is 6×103235Unuclei .

The composition of 238U from 4.5 billion years ago is calculated by the formula,

Composition of238U=Amount of 238U when earth was formedAmountof(238U+235U) whenearthwasformed×100 =1.9×1041.9×104+6×103×100=76%

The composition of 238U from 4.5 billion years ago is 76%_ .

The composition of 235U from 4.5 billion years ago is calculated by the formula,

Composition of235U=Amount of235U when earth was formedAmountof(238U+235U)whenearthwasformed×100 =6×1031.9×104+6×103×100=24%

The composition of 238U from 4.5 billion years ago is 24%_ .

Conclusion

Conclusion

The relative abundance of 238U when earth was formed is 76%_ .

The relative abundance of 235U when earth was formed is 24%_ .

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Chapter 18 Solutions

Chemistry: An Atoms First Approach

Ch. 18 - Prob. 1QCh. 18 - Prob. 2QCh. 18 - Prob. 3QCh. 18 - Prob. 4QCh. 18 - Prob. 5QCh. 18 - Prob. 6QCh. 18 - Prob. 7QCh. 18 - Prob. 8QCh. 18 - Prob. 9QCh. 18 - Prob. 10QCh. 18 - Prob. 11ECh. 18 - Prob. 12ECh. 18 - Prob. 13ECh. 18 - Prob. 14ECh. 18 - Prob. 15ECh. 18 - Prob. 16ECh. 18 - Prob. 17ECh. 18 - Prob. 18ECh. 18 - Prob. 19ECh. 18 - Prob. 20ECh. 18 - Prob. 21ECh. 18 - Prob. 22ECh. 18 - Prob. 23ECh. 18 - Prob. 24ECh. 18 - Prob. 25ECh. 18 - Prob. 26ECh. 18 - Prob. 27ECh. 18 - Prob. 28ECh. 18 - Prob. 29ECh. 18 - Prob. 30ECh. 18 - Prob. 31ECh. 18 - Prob. 32ECh. 18 - Prob. 33ECh. 18 - Prob. 34ECh. 18 - Prob. 35ECh. 18 - Prob. 36ECh. 18 - Prob. 37ECh. 18 - Prob. 38ECh. 18 - Prob. 39ECh. 18 - Prob. 40ECh. 18 - Prob. 41ECh. 18 - Prob. 42ECh. 18 - Prob. 43ECh. 18 - Prob. 44ECh. 18 - Prob. 45ECh. 18 - Prob. 46ECh. 18 - Prob. 47ECh. 18 - Prob. 48ECh. 18 - Prob. 49ECh. 18 - Prob. 50ECh. 18 - Prob. 51ECh. 18 - Prob. 52ECh. 18 - Prob. 53ECh. 18 - A chemist studied the reaction mechanism for the...Ch. 18 - Prob. 55ECh. 18 - Prob. 56ECh. 18 - Prob. 57ECh. 18 - Prob. 58ECh. 18 - Prob. 59AECh. 18 - Prob. 60AECh. 18 - Prob. 61AECh. 18 - Prob. 62AECh. 18 - Prob. 63AECh. 18 - Prob. 64AECh. 18 - Prob. 65AECh. 18 - Prob. 66AECh. 18 - Prob. 67AECh. 18 - Prob. 68AECh. 18 - Prob. 69AECh. 18 - Prob. 70AECh. 18 - Prob. 71AECh. 18 - Prob. 72AECh. 18 - Prob. 73CWPCh. 18 - Prob. 74CWPCh. 18 - Prob. 75CWPCh. 18 - Prob. 76CWPCh. 18 - Prob. 77CWPCh. 18 - Prob. 78CWPCh. 18 - Prob. 79CPCh. 18 - Prob. 80CPCh. 18 - Prob. 81CPCh. 18 - Prob. 82CPCh. 18 - Prob. 83CPCh. 18 - Prob. 84CPCh. 18 - Prob. 85CPCh. 18 - Prob. 86CPCh. 18 - Prob. 87IPCh. 18 - Prob. 88IP
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