College Physics:
College Physics:
11th Edition
ISBN: 9781305965515
Author: SERWAY, Raymond A.
Publisher: Brooks/Cole Pub Co
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Chapter 29, Problem 26P

(a)

To determine

The number of C137 sample.

(a)

Expert Solution
Check Mark

Answer to Problem 26P

The number of C137 sample is 3.54×109 nuclei.

Explanation of Solution

Given info: Time elapsed is 1730 days. Half-life of C137 is 1.10×104days . Half-life of C134 is 734 days. The combined activity of both the samples is 11.0 Bq.

Formula to calculate the combined activity is,

R=(N0t1/2137)exp[(tt1/2137)ln2]+(N0t1/2134)exp[(tt1/2134)ln2]       (I)

  • t1/2137 is the half-life of C137 .
  • t1/2134 is the half-life of C134 .
  • N0 is the initial amount of sample.
  • t is the time elapsed.

From Equation (I), the initial amount of sample is,

N0=R(1t1/2137)exp[(tt1/2137)ln2]+(1t1/2134)exp[(tt1/2134)ln2]       (II)

Formula to calculate the number of C137 sample is,

N=N0exp[(tt1/2137)ln2]       (III)

Substitute Equation (II) in (III).

N=Rexp[(tt1/2137)ln2](1t1/2137)exp[(tt1/2137)ln2]+(1t1/2134)exp[(tt1/2134)ln2]

Substitute 11.0 Bq for R, 1.10×104days for t1/2137 , 734 days for t1/2134 and 1730 days for t in the above equation to get N.

N=(11.0Bq)exp[(1730days1.10×104days)ln2](11.10×104days)exp[(1730days1.10×104days)ln2]+(1734days)exp[(1730days734days)ln2]=3.54×109

Conclusion:

The number of C137 sample is 3.54×109 nuclei.

(b)

To determine

The number of C134 sample.

(b)

Expert Solution
Check Mark

Answer to Problem 26P

The number of C134 sample is 7.72×108 nuclei.

Explanation of Solution

Given info: Time elapsed is 1730 days. Half-life of C137 is 1.10×104days . Half-life of C134 is 734 days. The combined activity of both the samples is 11.0 Bq.

From Equation (II) of (a),

N0=R(1t1/2137)exp[(tt1/2137)ln2]+(1t1/2134)exp[(tt1/2134)ln2]

Formula to calculate the number of C134 sample is,

N=N0exp[(tt1/2134)ln2]       (IV)

Substitute Equation (II) in (IV).

N=Rexp[(tt1/2134)ln2](1t1/2137)exp[(tt1/2137)ln2]+(1t1/2134)exp[(tt1/2134)ln2]

Substitute 11.0 Bq for R, 1.10×104days for t1/2137 , 734 days for t1/2134 and 1730 days for t in the above equation to get N.

N=(11.0Bq)exp[(1730days734days)ln2](11.10×104days)exp[(1730days1.10×104days)ln2]+(1734days)exp[(1730days734days)ln2]=7.72×108

Conclusion:

The number of C134 sample is 7.72×108 nuclei.

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Chapter 29 Solutions

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