The number of 26 Al nuclei decay per second in a 1 mol sample of 26 Al should be explained. Concept Introduction: All radioactive decays show first order kinetics. This, half-life and decay constant are related to each other as follows: k = 0.693 t 1 2 Where, t 1 2 is half-life k is the decay constant

Question

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Answer 1

Question

Chapter 14, Problem 14.102PAE

(a)

Interpretation Introduction

Interpretation: The number of 26Al nuclei decay per second in a 1 mol sample of 26Al should be explained.

Concept Introduction:

All radioactive decays show first order kinetics. This, half-life and decay constant are related to each other as follows:

k=0.693t12

Where, t12 is half-life

k is the decay constant

(a)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.063×10−14 decays per second

Explanation of Solution

Calculating decay constant for A26l as follows:

k=0.693t 1 2 k=0.6937.17× 105 yr=9.66×10−7 yr−1

Now, calculate the number of decays per second in 1 mole of the A26l

k=9.66×10−7 yr−1( 1 yr 365 day)( 1 day 24 h)( 1 h 60 min)( 1 min 60 s)=3.063×10−14 s−1

(b)

Interpretation Introduction

Interpretation: Energy released in a single 26Al decay event should be calculated.

Concept Introduction:

The mass defect is calculated as the difference of masses of products and reactants. The energy released can be related to mass defect as follows:

E=Δmc2

Here, Δm is mass defect and c is speed of light.

Also, 1 amu = 931.7 MeV

(b)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.49 MeV

Explanation of Solution

Calculate the mass defect as follows:

Mass Defect

=|(Mass of M 26g+Mass of positron)−Mass of A26l|=|(25.982593+0.0005486)−25.986892|=3.75×10−3 amu

Since, 1amu =931.7 MeV

Thus,

3.75×10−3 amu=3.75×10−3×931.7 MeV=3.5 MeV

Thus, energy released will be 3.5 MeV.

(c)

Interpretation Introduction

Interpretation:

The heat evolved in one minute by the decay of the 1 mol sample of ²⁶Al from part (a) should be calculated.

Concept Introduction: In one mole of a substance there are 6.023×1023 number of particles.

(c)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.86×1012MeV

Explanation of Solution

Converting decay per second to decay per minute as follows:

k=3.063×10−14 s−1( 60 s 1 min)=1.84×10−12 min−1

In 1 mole there are 6.023×1023

number of particles also the calculated energy released is equal to heat thus, heat evolved per minute per mole will be:

Heat=3.5 MeV×6.023×1023 mol−1×1.837×10−12 min−1=3.87×1012 MeV mol−1 min−1

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Answer 2

Question

Chapter 14, Problem 14.102PAE

(a)

Interpretation Introduction

Interpretation: The number of 26Al nuclei decay per second in a 1 mol sample of 26Al should be explained.

Concept Introduction:

All radioactive decays show first order kinetics. This, half-life and decay constant are related to each other as follows:

k=0.693t12

Where, t12 is half-life

k is the decay constant

(a)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.063×10−14 decays per second

Explanation of Solution

Calculating decay constant for A26l as follows:

k=0.693t 1 2 k=0.6937.17× 105 yr=9.66×10−7 yr−1

Now, calculate the number of decays per second in 1 mole of the A26l

k=9.66×10−7 yr−1( 1 yr 365 day)( 1 day 24 h)( 1 h 60 min)( 1 min 60 s)=3.063×10−14 s−1

(b)

Interpretation Introduction

Interpretation: Energy released in a single 26Al decay event should be calculated.

Concept Introduction:

The mass defect is calculated as the difference of masses of products and reactants. The energy released can be related to mass defect as follows:

E=Δmc2

Here, Δm is mass defect and c is speed of light.

Also, 1 amu = 931.7 MeV

(b)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.49 MeV

Explanation of Solution

Calculate the mass defect as follows:

Mass Defect

=|(Mass of M 26g+Mass of positron)−Mass of A26l|=|(25.982593+0.0005486)−25.986892|=3.75×10−3 amu

Since, 1amu =931.7 MeV

Thus,

3.75×10−3 amu=3.75×10−3×931.7 MeV=3.5 MeV

Thus, energy released will be 3.5 MeV.

(c)

Interpretation Introduction

Interpretation:

The heat evolved in one minute by the decay of the 1 mol sample of ²⁶Al from part (a) should be calculated.

Concept Introduction: In one mole of a substance there are 6.023×1023 number of particles.

(c)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.86×1012MeV

Explanation of Solution

Converting decay per second to decay per minute as follows:

k=3.063×10−14 s−1( 60 s 1 min)=1.84×10−12 min−1

In 1 mole there are 6.023×1023

number of particles also the calculated energy released is equal to heat thus, heat evolved per minute per mole will be:

Heat=3.5 MeV×6.023×1023 mol−1×1.837×10−12 min−1=3.87×1012 MeV mol−1 min−1

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Answer 3

Question

Chapter 14, Problem 14.102PAE

(a)

Interpretation Introduction

Interpretation: The number of 26Al nuclei decay per second in a 1 mol sample of 26Al should be explained.

Concept Introduction:

All radioactive decays show first order kinetics. This, half-life and decay constant are related to each other as follows:

k=0.693t12

Where, t12 is half-life

k is the decay constant

(a)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.063×10−14 decays per second

Explanation of Solution

Calculating decay constant for A26l as follows:

k=0.693t 1 2 k=0.6937.17× 105 yr=9.66×10−7 yr−1

Now, calculate the number of decays per second in 1 mole of the A26l

k=9.66×10−7 yr−1( 1 yr 365 day)( 1 day 24 h)( 1 h 60 min)( 1 min 60 s)=3.063×10−14 s−1

(b)

Interpretation Introduction

Interpretation: Energy released in a single 26Al decay event should be calculated.

Concept Introduction:

The mass defect is calculated as the difference of masses of products and reactants. The energy released can be related to mass defect as follows:

E=Δmc2

Here, Δm is mass defect and c is speed of light.

Also, 1 amu = 931.7 MeV

(b)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.49 MeV

Explanation of Solution

Calculate the mass defect as follows:

Mass Defect

=|(Mass of M 26g+Mass of positron)−Mass of A26l|=|(25.982593+0.0005486)−25.986892|=3.75×10−3 amu

Since, 1amu =931.7 MeV

Thus,

3.75×10−3 amu=3.75×10−3×931.7 MeV=3.5 MeV

Thus, energy released will be 3.5 MeV.

(c)

Interpretation Introduction

Interpretation:

The heat evolved in one minute by the decay of the 1 mol sample of ²⁶Al from part (a) should be calculated.

Concept Introduction: In one mole of a substance there are 6.023×1023 number of particles.

(c)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.86×1012MeV

Explanation of Solution

Converting decay per second to decay per minute as follows:

k=3.063×10−14 s−1( 60 s 1 min)=1.84×10−12 min−1

In 1 mole there are 6.023×1023

number of particles also the calculated energy released is equal to heat thus, heat evolved per minute per mole will be:

Heat=3.5 MeV×6.023×1023 mol−1×1.837×10−12 min−1=3.87×1012 MeV mol−1 min−1

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Answer 4

Question

Chapter 14, Problem 14.102PAE

(a)

Interpretation Introduction

Interpretation: The number of 26Al nuclei decay per second in a 1 mol sample of 26Al should be explained.

Concept Introduction:

All radioactive decays show first order kinetics. This, half-life and decay constant are related to each other as follows:

k=0.693t12

Where, t12 is half-life

k is the decay constant

(a)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.063×10−14 decays per second

Explanation of Solution

Calculating decay constant for A26l as follows:

k=0.693t 1 2 k=0.6937.17× 105 yr=9.66×10−7 yr−1

Now, calculate the number of decays per second in 1 mole of the A26l

k=9.66×10−7 yr−1( 1 yr 365 day)( 1 day 24 h)( 1 h 60 min)( 1 min 60 s)=3.063×10−14 s−1

(b)

Interpretation Introduction

Interpretation: Energy released in a single 26Al decay event should be calculated.

Concept Introduction:

The mass defect is calculated as the difference of masses of products and reactants. The energy released can be related to mass defect as follows:

E=Δmc2

Here, Δm is mass defect and c is speed of light.

Also, 1 amu = 931.7 MeV

(b)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.49 MeV

Explanation of Solution

Calculate the mass defect as follows:

Mass Defect

=|(Mass of M 26g+Mass of positron)−Mass of A26l|=|(25.982593+0.0005486)−25.986892|=3.75×10−3 amu

Since, 1amu =931.7 MeV

Thus,

3.75×10−3 amu=3.75×10−3×931.7 MeV=3.5 MeV

Thus, energy released will be 3.5 MeV.

(c)

Interpretation Introduction

Interpretation:

The heat evolved in one minute by the decay of the 1 mol sample of ²⁶Al from part (a) should be calculated.

Concept Introduction: In one mole of a substance there are 6.023×1023 number of particles.

(c)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.86×1012MeV

Explanation of Solution

Converting decay per second to decay per minute as follows:

k=3.063×10−14 s−1( 60 s 1 min)=1.84×10−12 min−1

In 1 mole there are 6.023×1023

number of particles also the calculated energy released is equal to heat thus, heat evolved per minute per mole will be:

Heat=3.5 MeV×6.023×1023 mol−1×1.837×10−12 min−1=3.87×1012 MeV mol−1 min−1

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Answer 5

Chapter 14, Problem 14.102PAE

(a)

Interpretation Introduction

Interpretation: The number of 26Al nuclei decay per second in a 1 mol sample of 26Al should be explained.

Concept Introduction:

All radioactive decays show first order kinetics. This, half-life and decay constant are related to each other as follows:

k=0.693t12

Where, t12 is half-life

k is the decay constant

(a)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.063×10−14 decays per second

Explanation of Solution

Calculating decay constant for A26l as follows:

k=0.693t 1 2 k=0.6937.17× 105 yr=9.66×10−7 yr−1

Now, calculate the number of decays per second in 1 mole of the A26l

k=9.66×10−7 yr−1( 1 yr 365 day)( 1 day 24 h)( 1 h 60 min)( 1 min 60 s)=3.063×10−14 s−1

(b)

Interpretation Introduction

Interpretation: Energy released in a single 26Al decay event should be calculated.

Concept Introduction:

The mass defect is calculated as the difference of masses of products and reactants. The energy released can be related to mass defect as follows:

E=Δmc2

Here, Δm is mass defect and c is speed of light.

Also, 1 amu = 931.7 MeV

(b)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.49 MeV

Explanation of Solution

Calculate the mass defect as follows:

Mass Defect

=|(Mass of M 26g+Mass of positron)−Mass of A26l|=|(25.982593+0.0005486)−25.986892|=3.75×10−3 amu

Since, 1amu =931.7 MeV

Thus,

3.75×10−3 amu=3.75×10−3×931.7 MeV=3.5 MeV

Thus, energy released will be 3.5 MeV.

(c)

Interpretation Introduction

Interpretation:

The heat evolved in one minute by the decay of the 1 mol sample of ²⁶Al from part (a) should be calculated.

Concept Introduction: In one mole of a substance there are 6.023×1023 number of particles.

(c)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.86×1012MeV

Explanation of Solution

Converting decay per second to decay per minute as follows:

k=3.063×10−14 s−1( 60 s 1 min)=1.84×10−12 min−1

In 1 mole there are 6.023×1023

number of particles also the calculated energy released is equal to heat thus, heat evolved per minute per mole will be:

Heat=3.5 MeV×6.023×1023 mol−1×1.837×10−12 min−1=3.87×1012 MeV mol−1 min−1

Answer 6

Chapter 14, Problem 14.102PAE

(a)

Interpretation Introduction

Interpretation: The number of 26Al nuclei decay per second in a 1 mol sample of 26Al should be explained.

Concept Introduction:

All radioactive decays show first order kinetics. This, half-life and decay constant are related to each other as follows:

k=0.693t12

Where, t12 is half-life

k is the decay constant

(a)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.063×10−14 decays per second

Explanation of Solution

Calculating decay constant for A26l as follows:

k=0.693t 1 2 k=0.6937.17× 105 yr=9.66×10−7 yr−1

Now, calculate the number of decays per second in 1 mole of the A26l

k=9.66×10−7 yr−1( 1 yr 365 day)( 1 day 24 h)( 1 h 60 min)( 1 min 60 s)=3.063×10−14 s−1

Answer 7

Chapter 14, Problem 14.102PAE

(a)

Interpretation Introduction

Interpretation: The number of 26Al nuclei decay per second in a 1 mol sample of 26Al should be explained.

Concept Introduction:

All radioactive decays show first order kinetics. This, half-life and decay constant are related to each other as follows:

k=0.693t12

Where, t12 is half-life

k is the decay constant

Answer 8

(a)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.063×10−14 decays per second

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculating decay constant for A26l as follows:

k=0.693t 1 2 k=0.6937.17× 105 yr=9.66×10−7 yr−1

Now, calculate the number of decays per second in 1 mole of the A26l

k=9.66×10−7 yr−1( 1 yr 365 day)( 1 day 24 h)( 1 h 60 min)( 1 min 60 s)=3.063×10−14 s−1

Answer 13

(b)

Interpretation Introduction

Interpretation: Energy released in a single 26Al decay event should be calculated.

Concept Introduction:

The mass defect is calculated as the difference of masses of products and reactants. The energy released can be related to mass defect as follows:

E=Δmc2

Here, Δm is mass defect and c is speed of light.

Also, 1 amu = 931.7 MeV

(b)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.49 MeV

Explanation of Solution

Calculate the mass defect as follows:

Mass Defect

=|(Mass of M 26g+Mass of positron)−Mass of A26l|=|(25.982593+0.0005486)−25.986892|=3.75×10−3 amu

Since, 1amu =931.7 MeV

Thus,

3.75×10−3 amu=3.75×10−3×931.7 MeV=3.5 MeV

Thus, energy released will be 3.5 MeV.

Answer 14

(b)

Interpretation Introduction

Interpretation: Energy released in a single 26Al decay event should be calculated.

Concept Introduction:

The mass defect is calculated as the difference of masses of products and reactants. The energy released can be related to mass defect as follows:

E=Δmc2

Here, Δm is mass defect and c is speed of light.

Also, 1 amu = 931.7 MeV

Answer 15

(b)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.49 MeV

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculate the mass defect as follows:

Mass Defect

=|(Mass of M 26g+Mass of positron)−Mass of A26l|=|(25.982593+0.0005486)−25.986892|=3.75×10−3 amu

Since, 1amu =931.7 MeV

Thus,

3.75×10−3 amu=3.75×10−3×931.7 MeV=3.5 MeV

Thus, energy released will be 3.5 MeV.

Answer 20

(c)

Interpretation Introduction

Interpretation:

The heat evolved in one minute by the decay of the 1 mol sample of ²⁶Al from part (a) should be calculated.

Concept Introduction: In one mole of a substance there are 6.023×1023 number of particles.

(c)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.86×1012MeV

Explanation of Solution

Converting decay per second to decay per minute as follows:

k=3.063×10−14 s−1( 60 s 1 min)=1.84×10−12 min−1

In 1 mole there are 6.023×1023

number of particles also the calculated energy released is equal to heat thus, heat evolved per minute per mole will be:

Heat=3.5 MeV×6.023×1023 mol−1×1.837×10−12 min−1=3.87×1012 MeV mol−1 min−1

Answer 21

(c)

Interpretation Introduction

Interpretation:

The heat evolved in one minute by the decay of the 1 mol sample of ²⁶Al from part (a) should be calculated.

Concept Introduction: In one mole of a substance there are 6.023×1023 number of particles.

Answer 22

(c)

Expert Solution

Answer to Problem 14.102PAE

Solution:

3.86×1012MeV

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Converting decay per second to decay per minute as follows:

k=3.063×10−14 s−1( 60 s 1 min)=1.84×10−12 min−1

In 1 mole there are 6.023×1023

number of particles also the calculated energy released is equal to heat thus, heat evolved per minute per mole will be:

Heat=3.5 MeV×6.023×1023 mol−1×1.837×10−12 min−1=3.87×1012 MeV mol−1 min−1

Answer 27

Ch. 14 - Prob. 1CO Ch. 14 - Prob. 2CO Ch. 14 - Prob. 3CO Ch. 14 - Prob. 4CO Ch. 14 - Prob. 5CO Ch. 14 - Prob. 6CO Ch. 14 - Prob. 7CO Ch. 14 - Prob. 8CO Ch. 14 - Prob. 9CO Ch. 14 - Prob. 10CO

Concept explainers

Answer to Problem 14.102PAE

Explanation of Solution

Answer to Problem 14.102PAE

Explanation of Solution

Interpretation:

The heat evolved in one minute by the decay of the 1 mol sample of ²⁶Al from part (a) should be calculated.

Concept Introduction: In one mole of a substance there are 6.023×1023 number of particles.

Answer to Problem 14.102PAE

Explanation of Solution

Want to see more full solutions like this?

Chapter 14 Solutions

Concept explainers

Answer to Problem 14.102PAE

Explanation of Solution

Answer to Problem 14.102PAE

Explanation of Solution

Interpretation: The heat evolved in one minute by the decay of the 1 mol sample of 26Al from part (a) should be calculated. Concept Introduction: In one mole of a substance there are 6.023×1023 number of particles.

Answer to Problem 14.102PAE

Explanation of Solution

Want to see more full solutions like this?

Chapter 14 Solutions

Interpretation:

The heat evolved in one minute by the decay of the 1 mol sample of ²⁶Al from part (a) should be calculated.

Concept Introduction: In one mole of a substance there are 6.023×1023 number of particles.