The information regarding the treatment of a patient by BNCT technique that includes treating him with a compound that contains 10 B that gets accumulated inside cancer tumors and then the further irradiation with neutrons that are absorbed by 10 B nuclei and the formation of unstable form of 11 B undergoing alpha decay to 7 Li is given. Various questions based on the radioactivity involved in this treatment have to be answered. Concept introduction: The emission of radiation by unstable nuclei in the form of alpha, beta particles with a neutrino is called a radioactive decay. To determine: The balanced equation for the neutron absorption by 10 B and further alpha decay process.

Question

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Answer 1

Question

Chapter 19, Problem 19.69QP

(a)

Interpretation Introduction

Interpretation: The information regarding the treatment of a patient by BNCT technique that includes treating him with a compound that contains $10B$ that gets accumulated inside cancer tumors and then the further irradiation with neutrons that are absorbed by $10B$ nuclei and the formation of unstable form of $11B$ undergoing alpha decay to $7Li$ is given. Various questions based on the radioactivity involved in this treatment have to be answered.

Concept introduction: The emission of radiation by unstable nuclei in the form of alpha, beta particles with a neutrino is called a radioactive decay.

To determine: The balanced equation for the neutron absorption by $10B$ and further alpha decay process.

(a)

Expert Solution

Answer to Problem 19.69QP

Solution

The overall equation is given as,

$510B+01n→37Li+24He$

Explanation of Solution

Explanation

The equation representing the absorption of neutron by $10B$ is given as,

$510B+01n→511B$

The equation for the alpha decay in $11B$ is given as,

$511B→37Li+24He$

The overall equation is given as,

$510B+01n→37Li+24He$

(b)

Interpretation Introduction

To determine: The energy released by each nucleus of Boron $−10$ in the obtained reaction.

(b)

Expert Solution

Answer to Problem 19.69QP

Solution

The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

Explanation of Solution

Explanation

Given

The mass of $10B$ is $10.0129 amu$ .

The mass of $1n$ is $1.00866 amu$ .

The mass of $7Li$ is $7.01600 amu$ .

The mass of $4He$ is $4.00260 amu$ .

The reaction is,

$510B+01n→37Li+24He$

The formula for the amount of energy released during fusion is given as,

$ΔE=Δmc2$ (1)

Where,

$Δm$ is the change in the mass.
$c$ is the speed of the light $(3.0×108 m/s)$ .

The change in the mass is calculated as,

$Δm=Mass of products−Mass of reactants$

Substitute the mass of reactants and products in the above equation as,

$Δm=Mass of products−Mass of reactants=(7.01600 amu+4.00260 amu)−(10.0129 amu+1.00866 amu)=11.0186 amu−11.02156 amu=−0.00296 amu$

The conversion of $amu$ into $kg$ is done as,

$1 amu=1.66054×10−27 kg$

Hence, the conversion of $−0.00296 amu$ into $kg$ is done as,

$−0.00296 amu=−0.00296×1.66054×10−27 kg=−0.0049151984×10−27 kg$

Substitute the value of $Δm$ and $c$ in the equation (1),

$ΔE=Δmc2=−0.0049151984×10−27 kg×(3×108 m/s)2=−4.42×10−13 kg m2s−2$

The conversion of $kg m2s−2$ to $J$ is done as,

$1 kg m2s−2=1 J$

Hence, the conversion of $−4.42×10−13 kg m2s−2$ to $J$ is done as,

$−4.42×10−13 kg m2s−2=−4.42×10−13×1 J=-4.42×10-13 J_$

Therefore, energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

(c)

Interpretation Introduction

To determine: The reason behind the effective use of nuclide that undergoes alpha decay in the cancer therapy.

(c)

Expert Solution

Answer to Problem 19.69QP

Solution

The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy.

Explanation of Solution

Explanation

The nuclide that undergoes alpha decay is effective in cancer therapy because they have a high value of RBE and the second reason is that if a radionuclide is placed is placed inside a tumor then these alpha particles do not show any penetration into the healthy tissue.

Conclusion

a. The overall equation is given as,
$510B+01n→37Li+24He$
b. The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .
c. The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy

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Answer 2

Question

Chapter 19, Problem 19.69QP

(a)

Interpretation Introduction

Interpretation: The information regarding the treatment of a patient by BNCT technique that includes treating him with a compound that contains $10B$ that gets accumulated inside cancer tumors and then the further irradiation with neutrons that are absorbed by $10B$ nuclei and the formation of unstable form of $11B$ undergoing alpha decay to $7Li$ is given. Various questions based on the radioactivity involved in this treatment have to be answered.

Concept introduction: The emission of radiation by unstable nuclei in the form of alpha, beta particles with a neutrino is called a radioactive decay.

To determine: The balanced equation for the neutron absorption by $10B$ and further alpha decay process.

(a)

Expert Solution

Answer to Problem 19.69QP

Solution

The overall equation is given as,

$510B+01n→37Li+24He$

Explanation of Solution

Explanation

The equation representing the absorption of neutron by $10B$ is given as,

$510B+01n→511B$

The equation for the alpha decay in $11B$ is given as,

$511B→37Li+24He$

The overall equation is given as,

$510B+01n→37Li+24He$

(b)

Interpretation Introduction

To determine: The energy released by each nucleus of Boron $−10$ in the obtained reaction.

(b)

Expert Solution

Answer to Problem 19.69QP

Solution

The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

Explanation of Solution

Explanation

Given

The mass of $10B$ is $10.0129 amu$ .

The mass of $1n$ is $1.00866 amu$ .

The mass of $7Li$ is $7.01600 amu$ .

The mass of $4He$ is $4.00260 amu$ .

The reaction is,

$510B+01n→37Li+24He$

The formula for the amount of energy released during fusion is given as,

$ΔE=Δmc2$ (1)

Where,

$Δm$ is the change in the mass.
$c$ is the speed of the light $(3.0×108 m/s)$ .

The change in the mass is calculated as,

$Δm=Mass of products−Mass of reactants$

Substitute the mass of reactants and products in the above equation as,

$Δm=Mass of products−Mass of reactants=(7.01600 amu+4.00260 amu)−(10.0129 amu+1.00866 amu)=11.0186 amu−11.02156 amu=−0.00296 amu$

The conversion of $amu$ into $kg$ is done as,

$1 amu=1.66054×10−27 kg$

Hence, the conversion of $−0.00296 amu$ into $kg$ is done as,

$−0.00296 amu=−0.00296×1.66054×10−27 kg=−0.0049151984×10−27 kg$

Substitute the value of $Δm$ and $c$ in the equation (1),

$ΔE=Δmc2=−0.0049151984×10−27 kg×(3×108 m/s)2=−4.42×10−13 kg m2s−2$

The conversion of $kg m2s−2$ to $J$ is done as,

$1 kg m2s−2=1 J$

Hence, the conversion of $−4.42×10−13 kg m2s−2$ to $J$ is done as,

$−4.42×10−13 kg m2s−2=−4.42×10−13×1 J=-4.42×10-13 J_$

Therefore, energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

(c)

Interpretation Introduction

To determine: The reason behind the effective use of nuclide that undergoes alpha decay in the cancer therapy.

(c)

Expert Solution

Answer to Problem 19.69QP

Solution

The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy.

Explanation of Solution

Explanation

The nuclide that undergoes alpha decay is effective in cancer therapy because they have a high value of RBE and the second reason is that if a radionuclide is placed is placed inside a tumor then these alpha particles do not show any penetration into the healthy tissue.

Conclusion

a. The overall equation is given as,
$510B+01n→37Li+24He$
b. The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .
c. The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy

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Answer 3

Question

Chapter 19, Problem 19.69QP

(a)

Interpretation Introduction

Interpretation: The information regarding the treatment of a patient by BNCT technique that includes treating him with a compound that contains $10B$ that gets accumulated inside cancer tumors and then the further irradiation with neutrons that are absorbed by $10B$ nuclei and the formation of unstable form of $11B$ undergoing alpha decay to $7Li$ is given. Various questions based on the radioactivity involved in this treatment have to be answered.

Concept introduction: The emission of radiation by unstable nuclei in the form of alpha, beta particles with a neutrino is called a radioactive decay.

To determine: The balanced equation for the neutron absorption by $10B$ and further alpha decay process.

(a)

Expert Solution

Answer to Problem 19.69QP

Solution

The overall equation is given as,

$510B+01n→37Li+24He$

Explanation of Solution

Explanation

The equation representing the absorption of neutron by $10B$ is given as,

$510B+01n→511B$

The equation for the alpha decay in $11B$ is given as,

$511B→37Li+24He$

The overall equation is given as,

$510B+01n→37Li+24He$

(b)

Interpretation Introduction

To determine: The energy released by each nucleus of Boron $−10$ in the obtained reaction.

(b)

Expert Solution

Answer to Problem 19.69QP

Solution

The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

Explanation of Solution

Explanation

Given

The mass of $10B$ is $10.0129 amu$ .

The mass of $1n$ is $1.00866 amu$ .

The mass of $7Li$ is $7.01600 amu$ .

The mass of $4He$ is $4.00260 amu$ .

The reaction is,

$510B+01n→37Li+24He$

The formula for the amount of energy released during fusion is given as,

$ΔE=Δmc2$ (1)

Where,

$Δm$ is the change in the mass.
$c$ is the speed of the light $(3.0×108 m/s)$ .

The change in the mass is calculated as,

$Δm=Mass of products−Mass of reactants$

Substitute the mass of reactants and products in the above equation as,

$Δm=Mass of products−Mass of reactants=(7.01600 amu+4.00260 amu)−(10.0129 amu+1.00866 amu)=11.0186 amu−11.02156 amu=−0.00296 amu$

The conversion of $amu$ into $kg$ is done as,

$1 amu=1.66054×10−27 kg$

Hence, the conversion of $−0.00296 amu$ into $kg$ is done as,

$−0.00296 amu=−0.00296×1.66054×10−27 kg=−0.0049151984×10−27 kg$

Substitute the value of $Δm$ and $c$ in the equation (1),

$ΔE=Δmc2=−0.0049151984×10−27 kg×(3×108 m/s)2=−4.42×10−13 kg m2s−2$

The conversion of $kg m2s−2$ to $J$ is done as,

$1 kg m2s−2=1 J$

Hence, the conversion of $−4.42×10−13 kg m2s−2$ to $J$ is done as,

$−4.42×10−13 kg m2s−2=−4.42×10−13×1 J=-4.42×10-13 J_$

Therefore, energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

(c)

Interpretation Introduction

To determine: The reason behind the effective use of nuclide that undergoes alpha decay in the cancer therapy.

(c)

Expert Solution

Answer to Problem 19.69QP

Solution

The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy.

Explanation of Solution

Explanation

The nuclide that undergoes alpha decay is effective in cancer therapy because they have a high value of RBE and the second reason is that if a radionuclide is placed is placed inside a tumor then these alpha particles do not show any penetration into the healthy tissue.

Conclusion

a. The overall equation is given as,
$510B+01n→37Li+24He$
b. The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .
c. The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy

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Answer 4

Question

Chapter 19, Problem 19.69QP

(a)

Interpretation Introduction

Interpretation: The information regarding the treatment of a patient by BNCT technique that includes treating him with a compound that contains $10B$ that gets accumulated inside cancer tumors and then the further irradiation with neutrons that are absorbed by $10B$ nuclei and the formation of unstable form of $11B$ undergoing alpha decay to $7Li$ is given. Various questions based on the radioactivity involved in this treatment have to be answered.

Concept introduction: The emission of radiation by unstable nuclei in the form of alpha, beta particles with a neutrino is called a radioactive decay.

To determine: The balanced equation for the neutron absorption by $10B$ and further alpha decay process.

(a)

Expert Solution

Answer to Problem 19.69QP

Solution

The overall equation is given as,

$510B+01n→37Li+24He$

Explanation of Solution

Explanation

The equation representing the absorption of neutron by $10B$ is given as,

$510B+01n→511B$

The equation for the alpha decay in $11B$ is given as,

$511B→37Li+24He$

The overall equation is given as,

$510B+01n→37Li+24He$

(b)

Interpretation Introduction

To determine: The energy released by each nucleus of Boron $−10$ in the obtained reaction.

(b)

Expert Solution

Answer to Problem 19.69QP

Solution

The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

Explanation of Solution

Explanation

Given

The mass of $10B$ is $10.0129 amu$ .

The mass of $1n$ is $1.00866 amu$ .

The mass of $7Li$ is $7.01600 amu$ .

The mass of $4He$ is $4.00260 amu$ .

The reaction is,

$510B+01n→37Li+24He$

The formula for the amount of energy released during fusion is given as,

$ΔE=Δmc2$ (1)

Where,

$Δm$ is the change in the mass.
$c$ is the speed of the light $(3.0×108 m/s)$ .

The change in the mass is calculated as,

$Δm=Mass of products−Mass of reactants$

Substitute the mass of reactants and products in the above equation as,

$Δm=Mass of products−Mass of reactants=(7.01600 amu+4.00260 amu)−(10.0129 amu+1.00866 amu)=11.0186 amu−11.02156 amu=−0.00296 amu$

The conversion of $amu$ into $kg$ is done as,

$1 amu=1.66054×10−27 kg$

Hence, the conversion of $−0.00296 amu$ into $kg$ is done as,

$−0.00296 amu=−0.00296×1.66054×10−27 kg=−0.0049151984×10−27 kg$

Substitute the value of $Δm$ and $c$ in the equation (1),

$ΔE=Δmc2=−0.0049151984×10−27 kg×(3×108 m/s)2=−4.42×10−13 kg m2s−2$

The conversion of $kg m2s−2$ to $J$ is done as,

$1 kg m2s−2=1 J$

Hence, the conversion of $−4.42×10−13 kg m2s−2$ to $J$ is done as,

$−4.42×10−13 kg m2s−2=−4.42×10−13×1 J=-4.42×10-13 J_$

Therefore, energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

(c)

Interpretation Introduction

To determine: The reason behind the effective use of nuclide that undergoes alpha decay in the cancer therapy.

(c)

Expert Solution

Answer to Problem 19.69QP

Solution

The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy.

Explanation of Solution

Explanation

The nuclide that undergoes alpha decay is effective in cancer therapy because they have a high value of RBE and the second reason is that if a radionuclide is placed is placed inside a tumor then these alpha particles do not show any penetration into the healthy tissue.

Conclusion

a. The overall equation is given as,
$510B+01n→37Li+24He$
b. The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .
c. The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy

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Answer 5

Chapter 19, Problem 19.69QP

(a)

Interpretation Introduction

Interpretation: The information regarding the treatment of a patient by BNCT technique that includes treating him with a compound that contains $10B$ that gets accumulated inside cancer tumors and then the further irradiation with neutrons that are absorbed by $10B$ nuclei and the formation of unstable form of $11B$ undergoing alpha decay to $7Li$ is given. Various questions based on the radioactivity involved in this treatment have to be answered.

Concept introduction: The emission of radiation by unstable nuclei in the form of alpha, beta particles with a neutrino is called a radioactive decay.

To determine: The balanced equation for the neutron absorption by $10B$ and further alpha decay process.

(a)

Expert Solution

Answer to Problem 19.69QP

Solution

The overall equation is given as,

$510B+01n→37Li+24He$

Explanation of Solution

Explanation

The equation representing the absorption of neutron by $10B$ is given as,

$510B+01n→511B$

The equation for the alpha decay in $11B$ is given as,

$511B→37Li+24He$

The overall equation is given as,

$510B+01n→37Li+24He$

(b)

Interpretation Introduction

To determine: The energy released by each nucleus of Boron $−10$ in the obtained reaction.

(b)

Expert Solution

Answer to Problem 19.69QP

Solution

The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

Explanation of Solution

Explanation

Given

The mass of $10B$ is $10.0129 amu$ .

The mass of $1n$ is $1.00866 amu$ .

The mass of $7Li$ is $7.01600 amu$ .

The mass of $4He$ is $4.00260 amu$ .

The reaction is,

$510B+01n→37Li+24He$

The formula for the amount of energy released during fusion is given as,

$ΔE=Δmc2$ (1)

Where,

$Δm$ is the change in the mass.
$c$ is the speed of the light $(3.0×108 m/s)$ .

The change in the mass is calculated as,

$Δm=Mass of products−Mass of reactants$

Substitute the mass of reactants and products in the above equation as,

$Δm=Mass of products−Mass of reactants=(7.01600 amu+4.00260 amu)−(10.0129 amu+1.00866 amu)=11.0186 amu−11.02156 amu=−0.00296 amu$

The conversion of $amu$ into $kg$ is done as,

$1 amu=1.66054×10−27 kg$

Hence, the conversion of $−0.00296 amu$ into $kg$ is done as,

$−0.00296 amu=−0.00296×1.66054×10−27 kg=−0.0049151984×10−27 kg$

Substitute the value of $Δm$ and $c$ in the equation (1),

$ΔE=Δmc2=−0.0049151984×10−27 kg×(3×108 m/s)2=−4.42×10−13 kg m2s−2$

The conversion of $kg m2s−2$ to $J$ is done as,

$1 kg m2s−2=1 J$

Hence, the conversion of $−4.42×10−13 kg m2s−2$ to $J$ is done as,

$−4.42×10−13 kg m2s−2=−4.42×10−13×1 J=-4.42×10-13 J_$

Therefore, energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

(c)

Interpretation Introduction

To determine: The reason behind the effective use of nuclide that undergoes alpha decay in the cancer therapy.

(c)

Expert Solution

Answer to Problem 19.69QP

Solution

The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy.

Explanation of Solution

Explanation

The nuclide that undergoes alpha decay is effective in cancer therapy because they have a high value of RBE and the second reason is that if a radionuclide is placed is placed inside a tumor then these alpha particles do not show any penetration into the healthy tissue.

Conclusion

a. The overall equation is given as,
$510B+01n→37Li+24He$
b. The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .
c. The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy

Answer 6

Chapter 19, Problem 19.69QP

(a)

Interpretation Introduction

Interpretation: The information regarding the treatment of a patient by BNCT technique that includes treating him with a compound that contains $10B$ that gets accumulated inside cancer tumors and then the further irradiation with neutrons that are absorbed by $10B$ nuclei and the formation of unstable form of $11B$ undergoing alpha decay to $7Li$ is given. Various questions based on the radioactivity involved in this treatment have to be answered.

Concept introduction: The emission of radiation by unstable nuclei in the form of alpha, beta particles with a neutrino is called a radioactive decay.

To determine: The balanced equation for the neutron absorption by $10B$ and further alpha decay process.

(a)

Expert Solution

Answer to Problem 19.69QP

Solution

The overall equation is given as,

$510B+01n→37Li+24He$

Explanation of Solution

Explanation

The equation representing the absorption of neutron by $10B$ is given as,

$510B+01n→511B$

The equation for the alpha decay in $11B$ is given as,

$511B→37Li+24He$

The overall equation is given as,

$510B+01n→37Li+24He$

Answer 7

Chapter 19, Problem 19.69QP

(a)

Interpretation Introduction

Interpretation: The information regarding the treatment of a patient by BNCT technique that includes treating him with a compound that contains $10B$ that gets accumulated inside cancer tumors and then the further irradiation with neutrons that are absorbed by $10B$ nuclei and the formation of unstable form of $11B$ undergoing alpha decay to $7Li$ is given. Various questions based on the radioactivity involved in this treatment have to be answered.

Concept introduction: The emission of radiation by unstable nuclei in the form of alpha, beta particles with a neutrino is called a radioactive decay.

To determine: The balanced equation for the neutron absorption by $10B$ and further alpha decay process.

Answer 8

(a)

Expert Solution

Answer to Problem 19.69QP

Solution

The overall equation is given as,

$510B+01n→37Li+24He$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Explanation

The equation representing the absorption of neutron by $10B$ is given as,

$510B+01n→511B$

The equation for the alpha decay in $11B$ is given as,

$511B→37Li+24He$

The overall equation is given as,

$510B+01n→37Li+24He$

Answer 13

(b)

Interpretation Introduction

To determine: The energy released by each nucleus of Boron $−10$ in the obtained reaction.

(b)

Expert Solution

Answer to Problem 19.69QP

Solution

The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

Explanation of Solution

Explanation

Given

The mass of $10B$ is $10.0129 amu$ .

The mass of $1n$ is $1.00866 amu$ .

The mass of $7Li$ is $7.01600 amu$ .

The mass of $4He$ is $4.00260 amu$ .

The reaction is,

$510B+01n→37Li+24He$

The formula for the amount of energy released during fusion is given as,

$ΔE=Δmc2$ (1)

Where,

$Δm$ is the change in the mass.
$c$ is the speed of the light $(3.0×108 m/s)$ .

The change in the mass is calculated as,

$Δm=Mass of products−Mass of reactants$

Substitute the mass of reactants and products in the above equation as,

$Δm=Mass of products−Mass of reactants=(7.01600 amu+4.00260 amu)−(10.0129 amu+1.00866 amu)=11.0186 amu−11.02156 amu=−0.00296 amu$

The conversion of $amu$ into $kg$ is done as,

$1 amu=1.66054×10−27 kg$

Hence, the conversion of $−0.00296 amu$ into $kg$ is done as,

$−0.00296 amu=−0.00296×1.66054×10−27 kg=−0.0049151984×10−27 kg$

Substitute the value of $Δm$ and $c$ in the equation (1),

$ΔE=Δmc2=−0.0049151984×10−27 kg×(3×108 m/s)2=−4.42×10−13 kg m2s−2$

The conversion of $kg m2s−2$ to $J$ is done as,

$1 kg m2s−2=1 J$

Hence, the conversion of $−4.42×10−13 kg m2s−2$ to $J$ is done as,

$−4.42×10−13 kg m2s−2=−4.42×10−13×1 J=-4.42×10-13 J_$

Therefore, energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

Answer 14

(b)

Interpretation Introduction

To determine: The energy released by each nucleus of Boron $−10$ in the obtained reaction.

Answer 15

(b)

Expert Solution

Answer to Problem 19.69QP

Solution

The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Explanation

Given

The mass of $10B$ is $10.0129 amu$ .

The mass of $1n$ is $1.00866 amu$ .

The mass of $7Li$ is $7.01600 amu$ .

The mass of $4He$ is $4.00260 amu$ .

The reaction is,

$510B+01n→37Li+24He$

The formula for the amount of energy released during fusion is given as,

$ΔE=Δmc2$ (1)

Where,

$Δm$ is the change in the mass.
$c$ is the speed of the light $(3.0×108 m/s)$ .

The change in the mass is calculated as,

$Δm=Mass of products−Mass of reactants$

Substitute the mass of reactants and products in the above equation as,

$Δm=Mass of products−Mass of reactants=(7.01600 amu+4.00260 amu)−(10.0129 amu+1.00866 amu)=11.0186 amu−11.02156 amu=−0.00296 amu$

The conversion of $amu$ into $kg$ is done as,

$1 amu=1.66054×10−27 kg$

Hence, the conversion of $−0.00296 amu$ into $kg$ is done as,

$−0.00296 amu=−0.00296×1.66054×10−27 kg=−0.0049151984×10−27 kg$

Substitute the value of $Δm$ and $c$ in the equation (1),

$ΔE=Δmc2=−0.0049151984×10−27 kg×(3×108 m/s)2=−4.42×10−13 kg m2s−2$

The conversion of $kg m2s−2$ to $J$ is done as,

$1 kg m2s−2=1 J$

Hence, the conversion of $−4.42×10−13 kg m2s−2$ to $J$ is done as,

$−4.42×10−13 kg m2s−2=−4.42×10−13×1 J=-4.42×10-13 J_$

Therefore, energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .

Answer 20

(c)

Interpretation Introduction

To determine: The reason behind the effective use of nuclide that undergoes alpha decay in the cancer therapy.

(c)

Expert Solution

Answer to Problem 19.69QP

Solution

The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy.

Explanation of Solution

Explanation

The nuclide that undergoes alpha decay is effective in cancer therapy because they have a high value of RBE and the second reason is that if a radionuclide is placed is placed inside a tumor then these alpha particles do not show any penetration into the healthy tissue.

Conclusion

a. The overall equation is given as,
$510B+01n→37Li+24He$
b. The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .
c. The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy

Answer 21

(c)

Interpretation Introduction

To determine: The reason behind the effective use of nuclide that undergoes alpha decay in the cancer therapy.

Answer 22

(c)

Expert Solution

Answer to Problem 19.69QP

Solution

The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Explanation

The nuclide that undergoes alpha decay is effective in cancer therapy because they have a high value of RBE and the second reason is that if a radionuclide is placed is placed inside a tumor then these alpha particles do not show any penetration into the healthy tissue.

Conclusion

a. The overall equation is given as,
$510B+01n→37Li+24He$
b. The energy released by each nucleus of Boron $−10$ is $-4.42×10-13 J_$ .
c. The negligible penetration of particles showing alpha decay inside the healthy tissue and high RBE value are the reasons for the use of nuclide that undergoes alpha decay is effective in cancer therapy