The decay constant for 14 C .

Question

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Answer 1

Question

Chapter 29, Problem 41P

(a)

To determine

The decay constant for $14C$ .

(a)

Expert Solution

Answer to Problem 41P

The decay constant for $14C$ is $3.83×10−12 s−1$ .

Explanation of Solution

The activity of $14C$ in a living sample is $0.25 Bq/g$ of carbon. The half-life of $14C$ is $5730 yr$

Write the formula for half-life

$T1/2=τln2$ (I)

Here, $T1/2$ is the half-life and $τ$ is the time constant.

Write the formula for the decay constant

$λ=1τ$ (II)

Here, $λ$ is the decay constant

Conclusion:

Substitute equation (I) in (II) and rewrite to find the $λ$

$λ=ln2T1/2$

Substitute $5730 yr$ for $T1/2$ in the above relation to find the value of $λ$ . [Note: $1 yr=3.156×107 s$

$λ=ln25730 yr×3.156×107 s/yrλ=3.83×10−12 s−1$

Thus, the decay constant for $14C$ is $3.83×10−12 s−1$ .

(b)

To determine

Calculate the number of $14C$ atoms in $1.00 g$ of carbon.

(b)

Expert Solution

Answer to Problem 41P

The number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

Explanation of Solution

One mole of carbon atom has $12.011 g$ and the relative abundance of $14C$ is $1.3×10−12$ . The Avogadro number $NA=6.022×1023 atoms/mol$ .

The number of $14C$ atoms in $1.00 g$ of carbon is the total number of nuclei in $1.00 g$ of carbon times the relative abundance of $14C$ .

$N=massmass per mol×NA×relative abundance$

Substitute $6.022×1023 atoms/mol$ for $NA$ , $1.3×10−12$ for the relative abundance, $1.00 g$ for the mass and $12.011 g/mol$ for mass per mol.

$N=1.00 g12.011 g/mol×6.022×1023 atoms/mol×1.3×10−12N=6.5×1010 atoms$

Thus, the number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

(c)

To determine

The activity of $14C$ per gram of carbon in a living sample.

(c)

Expert Solution

Answer to Problem 41P

The activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

Explanation of Solution

Write the formula for the activity

$R=λN$ (III)

Here, $λ$ is the decay constant and $N$ is the number of nuclei after time $t$ .

Conclusion:

Substitute $3.83×10−12 s−1$ for $λ$ and $6.5×1010 atoms$ for $N$ in equation (III) to find the value of $R$

$R=3.83×10−12 s−1×6.5×1010 atomsR=0.25 Bq$

Divide by $1.00 g$ on both sides to find the value of activity per gram

$R1.00 g=0.25 Bq1.00 gR1.00 g=0.25 Bq/g$

Thus, the activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

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Tutorial Exercise An air-filled spherical capacitor is constructed with an inner-shell radius of 6.95 cm and an outer-shell radius of 14.5 cm. (a) Calculate the capacitance of the device. (b) What potential difference between the spheres results in a 4.00-μC charge on the capacitor? Part 1 of 4 - Conceptualize Since the separation between the inner and outer shells is much larger than a typical electronic capacitor with separation on the order of 0.1 mm and capacitance in the microfarad range, we expect the capacitance of this spherical configuration to be on the order of picofarads. The potential difference should be sufficiently low to avoid sparking through the air that separates the shells. Part 2 of 4 - Categorize We will calculate the capacitance from the equation for a spherical shell capacitor. We will then calculate the voltage found from Q = CAV.

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Answer 2

Question

Chapter 29, Problem 41P

(a)

To determine

The decay constant for $14C$ .

(a)

Expert Solution

Answer to Problem 41P

The decay constant for $14C$ is $3.83×10−12 s−1$ .

Explanation of Solution

The activity of $14C$ in a living sample is $0.25 Bq/g$ of carbon. The half-life of $14C$ is $5730 yr$

Write the formula for half-life

$T1/2=τln2$ (I)

Here, $T1/2$ is the half-life and $τ$ is the time constant.

Write the formula for the decay constant

$λ=1τ$ (II)

Here, $λ$ is the decay constant

Conclusion:

Substitute equation (I) in (II) and rewrite to find the $λ$

$λ=ln2T1/2$

Substitute $5730 yr$ for $T1/2$ in the above relation to find the value of $λ$ . [Note: $1 yr=3.156×107 s$

$λ=ln25730 yr×3.156×107 s/yrλ=3.83×10−12 s−1$

Thus, the decay constant for $14C$ is $3.83×10−12 s−1$ .

(b)

To determine

Calculate the number of $14C$ atoms in $1.00 g$ of carbon.

(b)

Expert Solution

Answer to Problem 41P

The number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

Explanation of Solution

One mole of carbon atom has $12.011 g$ and the relative abundance of $14C$ is $1.3×10−12$ . The Avogadro number $NA=6.022×1023 atoms/mol$ .

The number of $14C$ atoms in $1.00 g$ of carbon is the total number of nuclei in $1.00 g$ of carbon times the relative abundance of $14C$ .

$N=massmass per mol×NA×relative abundance$

Substitute $6.022×1023 atoms/mol$ for $NA$ , $1.3×10−12$ for the relative abundance, $1.00 g$ for the mass and $12.011 g/mol$ for mass per mol.

$N=1.00 g12.011 g/mol×6.022×1023 atoms/mol×1.3×10−12N=6.5×1010 atoms$

Thus, the number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

(c)

To determine

The activity of $14C$ per gram of carbon in a living sample.

(c)

Expert Solution

Answer to Problem 41P

The activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

Explanation of Solution

Write the formula for the activity

$R=λN$ (III)

Here, $λ$ is the decay constant and $N$ is the number of nuclei after time $t$ .

Conclusion:

Substitute $3.83×10−12 s−1$ for $λ$ and $6.5×1010 atoms$ for $N$ in equation (III) to find the value of $R$

$R=3.83×10−12 s−1×6.5×1010 atomsR=0.25 Bq$

Divide by $1.00 g$ on both sides to find the value of activity per gram

$R1.00 g=0.25 Bq1.00 gR1.00 g=0.25 Bq/g$

Thus, the activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

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Answer 3

Question

Chapter 29, Problem 41P

(a)

To determine

The decay constant for $14C$ .

(a)

Expert Solution

Answer to Problem 41P

The decay constant for $14C$ is $3.83×10−12 s−1$ .

Explanation of Solution

The activity of $14C$ in a living sample is $0.25 Bq/g$ of carbon. The half-life of $14C$ is $5730 yr$

Write the formula for half-life

$T1/2=τln2$ (I)

Here, $T1/2$ is the half-life and $τ$ is the time constant.

Write the formula for the decay constant

$λ=1τ$ (II)

Here, $λ$ is the decay constant

Conclusion:

Substitute equation (I) in (II) and rewrite to find the $λ$

$λ=ln2T1/2$

Substitute $5730 yr$ for $T1/2$ in the above relation to find the value of $λ$ . [Note: $1 yr=3.156×107 s$

$λ=ln25730 yr×3.156×107 s/yrλ=3.83×10−12 s−1$

Thus, the decay constant for $14C$ is $3.83×10−12 s−1$ .

(b)

To determine

Calculate the number of $14C$ atoms in $1.00 g$ of carbon.

(b)

Expert Solution

Answer to Problem 41P

The number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

Explanation of Solution

One mole of carbon atom has $12.011 g$ and the relative abundance of $14C$ is $1.3×10−12$ . The Avogadro number $NA=6.022×1023 atoms/mol$ .

The number of $14C$ atoms in $1.00 g$ of carbon is the total number of nuclei in $1.00 g$ of carbon times the relative abundance of $14C$ .

$N=massmass per mol×NA×relative abundance$

Substitute $6.022×1023 atoms/mol$ for $NA$ , $1.3×10−12$ for the relative abundance, $1.00 g$ for the mass and $12.011 g/mol$ for mass per mol.

$N=1.00 g12.011 g/mol×6.022×1023 atoms/mol×1.3×10−12N=6.5×1010 atoms$

Thus, the number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

(c)

To determine

The activity of $14C$ per gram of carbon in a living sample.

(c)

Expert Solution

Answer to Problem 41P

The activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

Explanation of Solution

Write the formula for the activity

$R=λN$ (III)

Here, $λ$ is the decay constant and $N$ is the number of nuclei after time $t$ .

Conclusion:

Substitute $3.83×10−12 s−1$ for $λ$ and $6.5×1010 atoms$ for $N$ in equation (III) to find the value of $R$

$R=3.83×10−12 s−1×6.5×1010 atomsR=0.25 Bq$

Divide by $1.00 g$ on both sides to find the value of activity per gram

$R1.00 g=0.25 Bq1.00 gR1.00 g=0.25 Bq/g$

Thus, the activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

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Answer 4

Question

Chapter 29, Problem 41P

(a)

To determine

The decay constant for $14C$ .

(a)

Expert Solution

Answer to Problem 41P

The decay constant for $14C$ is $3.83×10−12 s−1$ .

Explanation of Solution

The activity of $14C$ in a living sample is $0.25 Bq/g$ of carbon. The half-life of $14C$ is $5730 yr$

Write the formula for half-life

$T1/2=τln2$ (I)

Here, $T1/2$ is the half-life and $τ$ is the time constant.

Write the formula for the decay constant

$λ=1τ$ (II)

Here, $λ$ is the decay constant

Conclusion:

Substitute equation (I) in (II) and rewrite to find the $λ$

$λ=ln2T1/2$

Substitute $5730 yr$ for $T1/2$ in the above relation to find the value of $λ$ . [Note: $1 yr=3.156×107 s$

$λ=ln25730 yr×3.156×107 s/yrλ=3.83×10−12 s−1$

Thus, the decay constant for $14C$ is $3.83×10−12 s−1$ .

(b)

To determine

Calculate the number of $14C$ atoms in $1.00 g$ of carbon.

(b)

Expert Solution

Answer to Problem 41P

The number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

Explanation of Solution

One mole of carbon atom has $12.011 g$ and the relative abundance of $14C$ is $1.3×10−12$ . The Avogadro number $NA=6.022×1023 atoms/mol$ .

The number of $14C$ atoms in $1.00 g$ of carbon is the total number of nuclei in $1.00 g$ of carbon times the relative abundance of $14C$ .

$N=massmass per mol×NA×relative abundance$

Substitute $6.022×1023 atoms/mol$ for $NA$ , $1.3×10−12$ for the relative abundance, $1.00 g$ for the mass and $12.011 g/mol$ for mass per mol.

$N=1.00 g12.011 g/mol×6.022×1023 atoms/mol×1.3×10−12N=6.5×1010 atoms$

Thus, the number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

(c)

To determine

The activity of $14C$ per gram of carbon in a living sample.

(c)

Expert Solution

Answer to Problem 41P

The activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

Explanation of Solution

Write the formula for the activity

$R=λN$ (III)

Here, $λ$ is the decay constant and $N$ is the number of nuclei after time $t$ .

Conclusion:

Substitute $3.83×10−12 s−1$ for $λ$ and $6.5×1010 atoms$ for $N$ in equation (III) to find the value of $R$

$R=3.83×10−12 s−1×6.5×1010 atomsR=0.25 Bq$

Divide by $1.00 g$ on both sides to find the value of activity per gram

$R1.00 g=0.25 Bq1.00 gR1.00 g=0.25 Bq/g$

Thus, the activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

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Answer 5

Chapter 29, Problem 41P

(a)

To determine

The decay constant for $14C$ .

(a)

Expert Solution

Answer to Problem 41P

The decay constant for $14C$ is $3.83×10−12 s−1$ .

Explanation of Solution

The activity of $14C$ in a living sample is $0.25 Bq/g$ of carbon. The half-life of $14C$ is $5730 yr$

Write the formula for half-life

$T1/2=τln2$ (I)

Here, $T1/2$ is the half-life and $τ$ is the time constant.

Write the formula for the decay constant

$λ=1τ$ (II)

Here, $λ$ is the decay constant

Conclusion:

Substitute equation (I) in (II) and rewrite to find the $λ$

$λ=ln2T1/2$

Substitute $5730 yr$ for $T1/2$ in the above relation to find the value of $λ$ . [Note: $1 yr=3.156×107 s$

$λ=ln25730 yr×3.156×107 s/yrλ=3.83×10−12 s−1$

Thus, the decay constant for $14C$ is $3.83×10−12 s−1$ .

(b)

To determine

Calculate the number of $14C$ atoms in $1.00 g$ of carbon.

(b)

Expert Solution

Answer to Problem 41P

The number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

Explanation of Solution

One mole of carbon atom has $12.011 g$ and the relative abundance of $14C$ is $1.3×10−12$ . The Avogadro number $NA=6.022×1023 atoms/mol$ .

The number of $14C$ atoms in $1.00 g$ of carbon is the total number of nuclei in $1.00 g$ of carbon times the relative abundance of $14C$ .

$N=massmass per mol×NA×relative abundance$

Substitute $6.022×1023 atoms/mol$ for $NA$ , $1.3×10−12$ for the relative abundance, $1.00 g$ for the mass and $12.011 g/mol$ for mass per mol.

$N=1.00 g12.011 g/mol×6.022×1023 atoms/mol×1.3×10−12N=6.5×1010 atoms$

Thus, the number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

(c)

To determine

The activity of $14C$ per gram of carbon in a living sample.

(c)

Expert Solution

Answer to Problem 41P

The activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

Explanation of Solution

Write the formula for the activity

$R=λN$ (III)

Here, $λ$ is the decay constant and $N$ is the number of nuclei after time $t$ .

Conclusion:

Substitute $3.83×10−12 s−1$ for $λ$ and $6.5×1010 atoms$ for $N$ in equation (III) to find the value of $R$

$R=3.83×10−12 s−1×6.5×1010 atomsR=0.25 Bq$

Divide by $1.00 g$ on both sides to find the value of activity per gram

$R1.00 g=0.25 Bq1.00 gR1.00 g=0.25 Bq/g$

Thus, the activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

Answer 6

Chapter 29, Problem 41P

(a)

To determine

The decay constant for $14C$ .

(a)

Expert Solution

Answer to Problem 41P

The decay constant for $14C$ is $3.83×10−12 s−1$ .

Explanation of Solution

The activity of $14C$ in a living sample is $0.25 Bq/g$ of carbon. The half-life of $14C$ is $5730 yr$

Write the formula for half-life

$T1/2=τln2$ (I)

Here, $T1/2$ is the half-life and $τ$ is the time constant.

Write the formula for the decay constant

$λ=1τ$ (II)

Here, $λ$ is the decay constant

Conclusion:

Substitute equation (I) in (II) and rewrite to find the $λ$

$λ=ln2T1/2$

Substitute $5730 yr$ for $T1/2$ in the above relation to find the value of $λ$ . [Note: $1 yr=3.156×107 s$

$λ=ln25730 yr×3.156×107 s/yrλ=3.83×10−12 s−1$

Thus, the decay constant for $14C$ is $3.83×10−12 s−1$ .

Answer 7

Chapter 29, Problem 41P

(a)

To determine

The decay constant for $14C$ .

Answer 8

(a)

Expert Solution

Answer to Problem 41P

The decay constant for $14C$ is $3.83×10−12 s−1$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The activity of $14C$ in a living sample is $0.25 Bq/g$ of carbon. The half-life of $14C$ is $5730 yr$

Write the formula for half-life

$T1/2=τln2$ (I)

Here, $T1/2$ is the half-life and $τ$ is the time constant.

Write the formula for the decay constant

$λ=1τ$ (II)

Here, $λ$ is the decay constant

Conclusion:

Substitute equation (I) in (II) and rewrite to find the $λ$

$λ=ln2T1/2$

Substitute $5730 yr$ for $T1/2$ in the above relation to find the value of $λ$ . [Note: $1 yr=3.156×107 s$

$λ=ln25730 yr×3.156×107 s/yrλ=3.83×10−12 s−1$

Thus, the decay constant for $14C$ is $3.83×10−12 s−1$ .

Answer 13

(b)

To determine

Calculate the number of $14C$ atoms in $1.00 g$ of carbon.

(b)

Expert Solution

Answer to Problem 41P

The number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

Explanation of Solution

One mole of carbon atom has $12.011 g$ and the relative abundance of $14C$ is $1.3×10−12$ . The Avogadro number $NA=6.022×1023 atoms/mol$ .

The number of $14C$ atoms in $1.00 g$ of carbon is the total number of nuclei in $1.00 g$ of carbon times the relative abundance of $14C$ .

$N=massmass per mol×NA×relative abundance$

Substitute $6.022×1023 atoms/mol$ for $NA$ , $1.3×10−12$ for the relative abundance, $1.00 g$ for the mass and $12.011 g/mol$ for mass per mol.

$N=1.00 g12.011 g/mol×6.022×1023 atoms/mol×1.3×10−12N=6.5×1010 atoms$

Thus, the number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

Answer 14

(b)

To determine

Calculate the number of $14C$ atoms in $1.00 g$ of carbon.

Answer 15

(b)

Expert Solution

Answer to Problem 41P

The number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

One mole of carbon atom has $12.011 g$ and the relative abundance of $14C$ is $1.3×10−12$ . The Avogadro number $NA=6.022×1023 atoms/mol$ .

The number of $14C$ atoms in $1.00 g$ of carbon is the total number of nuclei in $1.00 g$ of carbon times the relative abundance of $14C$ .

$N=massmass per mol×NA×relative abundance$

Substitute $6.022×1023 atoms/mol$ for $NA$ , $1.3×10−12$ for the relative abundance, $1.00 g$ for the mass and $12.011 g/mol$ for mass per mol.

$N=1.00 g12.011 g/mol×6.022×1023 atoms/mol×1.3×10−12N=6.5×1010 atoms$

Thus, the number of $14C$ atoms in $1.00 g$ of carbon is $6.5×1010 atoms$ .

Answer 20

(c)

To determine

The activity of $14C$ per gram of carbon in a living sample.

(c)

Expert Solution

Answer to Problem 41P

The activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

Explanation of Solution

Write the formula for the activity

$R=λN$ (III)

Here, $λ$ is the decay constant and $N$ is the number of nuclei after time $t$ .

Conclusion:

Substitute $3.83×10−12 s−1$ for $λ$ and $6.5×1010 atoms$ for $N$ in equation (III) to find the value of $R$

$R=3.83×10−12 s−1×6.5×1010 atomsR=0.25 Bq$

Divide by $1.00 g$ on both sides to find the value of activity per gram

$R1.00 g=0.25 Bq1.00 gR1.00 g=0.25 Bq/g$

Thus, the activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

Answer 21

(c)

To determine

The activity of $14C$ per gram of carbon in a living sample.

Answer 22

(c)

Expert Solution

Answer to Problem 41P

The activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the formula for the activity

$R=λN$ (III)

Here, $λ$ is the decay constant and $N$ is the number of nuclei after time $t$ .

Conclusion:

Substitute $3.83×10−12 s−1$ for $λ$ and $6.5×1010 atoms$ for $N$ in equation (III) to find the value of $R$

$R=3.83×10−12 s−1×6.5×1010 atomsR=0.25 Bq$

Divide by $1.00 g$ on both sides to find the value of activity per gram

$R1.00 g=0.25 Bq1.00 gR1.00 g=0.25 Bq/g$

Thus, the activity of $14C$ per gram of carbon in a living sample is $0.25 Bq/g$ .

Answer 27

Ch. 29.1 - Prob. 29.1CP Ch. 29.1 - Prob. 29.1PP Ch. 29.1 - Prob. 29.2PP Ch. 29.1 - Prob. 29.3PP Ch. 29.2 - Prob. 29.4PP Ch. 29.3 - Prob. 29.5PP Ch. 29.3 - Prob. 29.6PP Ch. 29.3 - Prob. 29.7PP Ch. 29.4 - Prob. 29.4CP Ch. 29.4 - Prob. 29.8PP

The decay constant for 14 C .

Concept explainers

The decay constant for $14C$ .

Answer to Problem 41P

Explanation of Solution

Calculate the number of $14C$ atoms in $1.00 g$ of carbon.

Answer to Problem 41P

Explanation of Solution

The activity of $14C$ per gram of carbon in a living sample.

Answer to Problem 41P

Explanation of Solution

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Chapter 29 Solutions

The decay constant for 14 C .

Concept explainers

The decay constant for 14C<m:math><m:mrow><m:msup><m:mrow></m:mrow><m:mrow><m:mn>14</m:mn></m:mrow></m:msup><m:mtext>C</m:mtext></m:mrow></m:math>.

Answer to Problem 41P

Explanation of Solution

Calculate the number of 14C<m:math><m:mrow><m:msup><m:mrow></m:mrow><m:mrow><m:mn>14</m:mn></m:mrow></m:msup><m:mtext>C</m:mtext></m:mrow></m:math> atoms in 1.00 g<m:math><m:mrow><m:mn>1.00</m:mn><m:mtext> g</m:mtext></m:mrow></m:math> of carbon.

Answer to Problem 41P

Explanation of Solution

The activity of 14C<m:math><m:mrow><m:msup><m:mrow></m:mrow><m:mrow><m:mn>14</m:mn></m:mrow></m:msup><m:mtext>C</m:mtext></m:mrow></m:math> per gram of carbon in a living sample.

Answer to Problem 41P

Explanation of Solution

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Chapter 29 Solutions

The decay constant for $14C$ .

Calculate the number of $14C$ atoms in $1.00 g$ of carbon.

The activity of $14C$ per gram of carbon in a living sample.