Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 43, Problem 63AP

(a)

To determine

The activity in curies due to each of the isotopes.

(a)

Expert Solution
Check Mark

Answer to Problem 63AP

The activity in curies due to each of the isotopes is 330μCi, 16μCi and 310μCi.

Explanation of Solution

Write the expression to find the mass of 238U in the sample.

  m238U=Mm235Um234U

Here, m238U is the mass of 238U, M is the total mass of the sample, m235U is the mass of 235U, m234U is the mass of 234U.

Substitute 1.00kg for M, (1.00 kg)(0.00720) for m235U and (1.00 kg)(0.0000500) for 234U to find the mass of 238U in the sample.

  m238U=1.00kg(1.00 kg)(0.00720)(1.00 kg)(0.0000500)=0.993kg

Conclusion:

Write the expression to find the number of nuclei in 238U.

  N=m238U(6.02×1023 nuclei mol)(1 mol0.238 kg)

Substitute 0.993kg for m238U to find the number of nuclei in 238U.

  N=0.993kg(6.02×1023 nuclei mol)(1 mol0.238 kg)=2.51×1024nuclei

Write the expression to find the activity of 238U

  R=λN=ln2T1/2N

Here, T1/2 is the half-life, λ is the decay constant.

Substitute 2.51×1024nuclei for N and 4.47×109yr for T1/2 to find the activity of 238U.

  R=ln24.47×109yr(2.51×1024nuclei)(1 yr3.16×107s)(1 Ci3.70×1010s1)=3.3×104Ci106μCi1.0Ci=330μCi

Write the expression to find the number of nuclei in 235U.

  N=m235U(6.02×1023 nuclei mol)(1 mol0.238 kg)

Substitute (1.00 kg)(0.00720) for m235U to find the number of nuclei in 235U.

  N=(1.00 kg)(0.00720)(6.02×1023 nuclei mol)(1 mol0.238 kg)=1.84×1022nuclei

Write the expression to find the activity of 235U

  R=λN=ln2T1/2N

Here, T1/2 is the half-life, λ is the decay constant.

Substitute 1.84×1022nuclei for N and 4.47×109yr for T1/2 to find the activity of 235U.

  R=ln27.04×108 yr(1.84×1022nuclei)(1 yr3.16×107s)(1 Ci3.70×1010s1)=3.6×105Ci106μCi1.0Ci=16μCi

Write the expression to find the number of nuclei in 234U.

  N=m234U(6.02×1023 nuclei mol)(1 mol0.238 kg)

Substitute (1.00kg)(0.0000500) for m234U to find the number of nuclei in 234U.

  N=(1.00kg)(0.0000500)(6.02×1023 nuclei mol)(1 mol0.238 kg)=1.29×1020nuclei

Write the expression to find the activity of 234U

  R=λN=ln2T1/2N

Here, T1/2 is the half-life, λ is the decay constant.

Substitute 1.29×1020nuclei for N and 4.47×109yr for T1/2 to find the activity of 234U.

  R=ln22.44×105 yr(1.29×1020nuclei)(1 yr3.16×107s)(1 Ci3.70×1010s1)=3.1×104Ci106μCi1.0Ci=310μCi

Thus, The activity in curies due to each of the isotopes is 330μCi, 16μCi and 310μCi.

(b)

To determine

The fraction of the total activity due to isotopes.

(b)

Expert Solution
Check Mark

Answer to Problem 63AP

The fraction of the total activity due to isotopes are 50%, 2.4%, 47%.

Explanation of Solution

the activity of 238U is 330μCi, the activity of 235U is 16μCi, the activity of 234U is 310μCi.

Write the expression to find the total activity of the sample.

  Rt=R238U+R235U+R234U

Here, Rt is the total activity of the sample, R238U is the activity of 238U, R235U is the activity of 235U and R234U is the activity of 234U.

Substitute 330μCi for R238U, 16μCi for R235U and 310μCi for R234U to find Rt.

  Rt=330μCi+16μCi+310μCi=656μCi

Conclusion:

Thus, the fraction of activity due to 238U is 330μCi656μCi×100%=50%, the fraction of activity due to 235U is 16μCi656μCi×100%=2.4%, the fraction of activity due to 234U is 310μCi656μCi×100%=47%.

(c)

To determine

Whether activity of the sample is dangerous.

(c)

Expert Solution
Check Mark

Answer to Problem 63AP

Yes. The activity of the sample is dangerous.

Explanation of Solution

The activity of 238U is 330μCi, the activity of 235U is 16μCi, the activity of 234U is 310μCi.

Write the expression to find the total activity of the sample.

  Rt=R238U+R235U+R234U

Here, Rt is the total activity of the sample, R238U is the activity of 238U, R235U is the activity of 235U and R234U is the activity of 234U.

Substitute 330μCi for R238U, 16μCi for R235U and 310μCi for R234U to find Rt.

  Rt=330μCi+16μCi+310μCi=656μCi

Conclusion:

Since, the activity of the sample is 656μCi; it is dangerous to human. But in the laboratories, it is used with taking enough precautions to reduce the human contact with the sample.

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