Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 30, Problem 24P

(a)

To determine

The number of carbon atoms in the given sample.

(a)

Expert Solution
Check Mark

Answer to Problem 24P

The number of carbon atoms in the given sample is 1.05×1021.

Explanation of Solution

Write the equation to find the number of carbon atoms.

    N=mMCNava

Here, N is the number of carbon atoms, m is the given mass in gram, MC is the atomic mass of carbon, and Nava is the Avogadro number.

Conclusion:

Substitute 0.021 0g for m, 12.0 g/mol for MC, 6.02×1023atoms/mol for Nana in the above equation to find N.

    N=0.021 0g12.0 g/mol(6.02×1023atoms/mol)=1.05×1021

Therefore, the number of carbon atoms in the given sample is 1.05×1021.

(b)

To determine

The number of carbon atoms in the given sample.

(b)

Expert Solution
Check Mark

Answer to Problem 24P

The number of carbon-14 atoms in the given sample is 1.37×109.

Explanation of Solution

Write the equation to find the number of carbon-14 atoms.

    (N0)C-14=NNstable

Here, (N0)C-14 is the number of carbon-14 atoms and Nstable is the number of stable atoms which one carbon-14 is found.

Conclusion:

Substitute 1.05×1021 for N and 7.70×1011 for Nstable in the above equation to find (N0)C-14.

    (N0)C-14=1.05×10217.70×1011=1.37×109

Therefore, the number of carbon-14 atoms in the given sample is 1.37×109.

(c)

To determine

The decay constant for carbon-14 in inverse seconds.

(c)

Expert Solution
Check Mark

Answer to Problem 24P

The decay constant for carbon-14 in inverse seconds is 3.83×1012 s1.

Explanation of Solution

Write the equation to find the half-life time of carbon-14.

    λC-14=ln2t1/2

Here, λC-14 is the decay constant and t1/2 is the half life time.

Conclusion:

Substitute 5730yr in the above equation to find λC-14.

    λC-14=ln25730yr(3.16×107s1yr)=3.83×1012 s1

Therefore, the decay constant for carbon-14 in inverse seconds is 3.83×1012 s1.

(d)

To determine

The number of initial number of decays in a week immediately after the death of species.

(d)

Expert Solution
Check Mark

Answer to Problem 24P

The number of initial number of decays in a week immediately after the death of species is 3.17×103decays/week.

Explanation of Solution

Write the equation to find the decay rate.

    R=λC-14N

Here, R is the number of decay and N is the number of C-14 atoms at time t.

Write the equation to find N.

  N=(N0)C-14eλt

Here, t is the time.

Rewrite the equation for R by substituting the above relation for N.

  R=λC-14((N0)C-14eλt)

To find the R at initial time, rewrite the above relation by substituting 0s for t.

  R=λC-14((N0)C-14eλ(0s))=λC-14(N0)C-14

Conclusion:

Substitute 3.83×1012s1 for λC-14 and 1.37×109 for (N0)C-14 in the above equation to find R

    R=(3.83×1012s1)(1.37×109(7(86 400 s)1week))=3.17×103decays/week

Therefore, the number of initial number of decays in a week immediately after the death of species is 3.17×103decays/week.

(e)

To determine

The new number of decays in a week in the current sample.

(e)

Expert Solution
Check Mark

Answer to Problem 24P

The new number of decays in a week in the current sample is 951 decays/week.

Explanation of Solution

Write the equation to find the new number of decays in a week in the current sample.

    R=ncountsη

Here, ncounts is the number of counts accumulated in a week and η is the counting efficiency.

Conclusion:

Substitute 837 for ncounts and 88% for η in the above equation to find R.

    R=837decays/week(88%100%)=837decays/week0.88=951decays/week

Therefore, the new number of decays in a week in the current sample is 951 decays/week.

(f)

To determine

Lifetime of specimen in years using the results from part (c) and (d).

(f)

Expert Solution
Check Mark

Answer to Problem 24P

Lifetime of specimen is 9.95×103yr.

Explanation of Solution

Write the equation to find the fraction of decay.

    RR0=eλt

Apply logarithm on both sides.

  lnRR0=lneλtlnRR0=λt

Rewrite the above equation in terms of t.

  t=1λln(RR0)

Conclusion:

Substitute 1.21×104 yr1 for λ, 951decays/week for R, and 3.17×103decays/week for R0 in the above equation to find t.

  t=1(1.21×104 yr1)ln(951decays/week3.17×103decays/week)=(8264.46yr)(ln0.3)=(8264.46yr)(1.20)=9.95×103yr

Therefore, the lifetime of specimen is 9.95×103yr.

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Chapter 30 Solutions

Principles of Physics: A Calculus-Based Text

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