General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
Question
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Chapter 30, Problem 39E

(a)

To determine

The number of 14C atoms present in 1kg wood.

(a)

Expert Solution
Check Mark

Answer to Problem 39E

The number of 14C atoms present in 1kg wood is 5.52×1013.

Explanation of Solution

Write the expression for the number of 12C atoms present in 1Kg wood.

    n=(NM)m        (I)

Here, N is the number of 12C nuclei per mole, M is the molar mass of 12C and m is the mass of the carbon.

Write the expression for the number of 14C present.

    n=n(pP)        (II)

Here, n is the number of 14C present, p is the number of 14C present for P number of 12C atoms.

Conclusion:

Substitute 6.02×1023 for N, 12gm for M and 1kg for m in expression (I).

    n=(6.02×102312gm)(1kg)(1000g1kg)=(6.623×102312gm)(1000gm)=5.01×1025

Substitute 1 for p, 1012 for P and 5.52×1022 for n in expression (II).

    n=(5.01×1025)(11012)=5.01×1013

Thus, the number of 14C atoms present in 1kg wood is 5.52×1013.

(b)

To determine

The number of the atoms that will decay in 1hr.

(b)

Expert Solution
Check Mark

Answer to Problem 39E

The number of the atoms that will decay in 1hr is 5.519×1013.

Explanation of Solution

Write the expression for the number of atoms that will decay.

    N=N0eλt        (II)

Here, N is the number of atom present after time t, N0 is the initial number of atom present and λ is the decay constant.

Conclusion:

Substitute 5.52×1010 for N0, 3.84×1012s1 for λ and 1hr for t in expression (II).

    N=(5.52×1013)e((3.84×1012s1)(1hr))=(5.52×1013)e((3.84×1012s1)(3600s))5.519×1013

Thus, the number of atoms that will decay in 1hr is 5.519×1013.

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