Chapter 24, Problem 24.114P

(a)

Interpretation Introduction

Interpretation:

Product formed by the fusion of two ${}^{6}Li$ nuclei has to be written.

Concept Introduction:

Nuclear fusion is the reaction between two or more nuclei and which comes close enough to form one or more different atomic nuclei and subatomic particle.

Explanation of Solution

Fusion of two ${}^{6}Li$ nuclei takes place. Atomic number of the product is 6 (Carbon) and mass number is 12.

Therefore, the reaction can be given as,

  ${}_3^{6}Li{\text{ + }}_3^{6}Li\underset{}{\to }{}_6^{12}C\left(\text{dilithium}\right)$

(b)

Interpretation Introduction

Interpretation:

Energy released during the reaction has to be determined.

Concept Introduction:

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  $\text{Nucleus + nuclear binding energy }\underset{}{\to }\text{ nucleons}$

It can be calculated using the given formula,

  $\begin{array}{l}\Delta \text{E = }\Delta {\text{mc}}^2\\ \\ where,\\ \Delta \text{m = Mass Difference}\\ c=\text{ Speed of light}\end{array}$

Change in mass of a given reaction can be determined as given,

  $\Delta m\text{ = }Mass\text{ of reactant}-\text{ Mass of products}$

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}Given{\text{ reaction: }}_3^{6}Li{\text{ + }}_3^{6}Li\underset{}{\to }{}_6^{12}C\\ \\ {\text{Mass of }}_3^{6}Li=\text{ 6}\text{.015121 amu}\\ \\ {\text{Mass of }}_6^{12}\text{C}=\text{ 12}\text{.000000 amu}\end{array}$

• Calculate the mass difference:

Mass difference of the reaction can be calculated as given,

  $\begin{array}{c}\Delta m\text{ = }Mass\text{ of reactant}-\text{ Mass of products}\\ \\ \text{= }\left[\text{2}\times Mass{\text{ of }}_3^{6}Li\right]-\left[{\text{ Mass of }}_6^{12}C\right]\\ \\ \text{= }\left[2\times \text{6}\text{.015121 amu}\right]-\left[\text{12}\text{.000000 amu}\right]\\ \\ =\text{ 0}{\text{.030242 amu/atom }}_6^{12}C\end{array}$

Mass difference is $\text{0}\text{.030242 amu/atom}\text{.}$

• Convert the unit of mass difference:

Unit of mass difference is converted from $\text{amu/atom}$ to $\text{kg/atom}$ as follows,

  $\begin{array}{c}\Delta m\text{ = }\left(\text{0}\text{.030242 amu/atom}\right)\left(1.66054\times {10}^{-27}\text{ kg/amu}\right)\\ \\ =\text{ 5}\text{.02180507}\times {10}^{-29}\text{ kg/atom}\end{array}$

• Calculate the energy per atom:

Energy per atom is calculated as follows for the reaction,

$\begin{array}{c}\text{Binding energy, E = }\Delta {\text{mc}}^2\\ \\ =\left(\text{5}\text{.02180507}\times {10}^{-29}\text{ kg/atom}\right){\left(2.998\times {10}^8\text{ m/s}\right)}^2\left(\frac{1\text{ J}}{1\text{ kg}{\text{.m}}^2.s^{-2}}\right)\\ \\ =\text{ 4}\text{.5133595}\times {\text{10}}^{-12}\text{ J/atom}\end{array}$

Energy per atom of the given reaction is $\text{4}\text{.5133595}\times {\text{10}}^{-12}\text{ J/atom}\text{.}$

• Convert the unit of energy:

Unit of energy is converted from $\text{J/atom}$ to $\text{J/kg}$ as given below,

  $\begin{array}{c}\text{E = }\left(\frac{\text{4}\text{.5133595}\times {\text{10}}^{-12}\text{ J}}{atom}\right)\left(\frac{1\text{ atom}}{\text{12}\text{.000000 amu}}\right)\left(\frac{1\text{ amu}}{1.66054\times {10}^{-27}\text{ kg}}\right)\\ \\ =\text{ 2}\text{.2650}\times {\text{10}}^{14}\text{ J/kg dilithium}\end{array}$

Therefore,

Energy released during the reaction is $2.2650\times 10^{14}J/kg.$

(c)

Interpretation Introduction

Interpretation:

Number of positron released during the given reaction has to be determined.

Concept Introduction:

Nuclear reaction: A nuclear reaction in which a lighter nucleus fuses together into new stable nuclei or a heavier nucleus split into stable daughter nuclei with the release of large amount of energy.

Common particles in radioactive decay and nuclear transformations are mentioned below,

  $\begin{array}{cc}\text{Particle}& \text{Symbol}\\ \text{Proton}& {}_{\text{1}}^{\text{1}}\text{H}\text{or}{}_{\text{1}}^{\text{1}}\text{P}\\ \text{Neutron}& {}_{\text{0}}^{\text{1}}\text{n}\\ \text{Electron}& {}_{\text{-1}}^{\text{0}}\text{e}\\ \text{Alpha}\text{particle}& {}_{\text{2}}^{\text{4}}\text{He}\text{or}{}_{\text{2}}^{\text{4}}\text{α}\\ \text{Beta}\text{particle}& {}_{\text{-1}}^{\text{0}}\text{e}\text{or}{}_{\text{-1}}^{\text{0}}\text{β}\\ \text{Positron}& {}_{\text{1}}^{\text{0}}\text{e}\end{array}$

Explanation of Solution

The given unbalanced nuclear equation is,

  $4{}_1^{1}H\underset{}{\to }{}_2^{4}He{\text{ + x }}_1^0\beta$

The positron particles do not have any impact on mass number. Hence, total atomic number released by the positron particles can be determined by taking the difference between the atomic number of Hydrogen and Helium.

$\begin{array}{c}\text{Total atomic no}\text{. for }{\beta }^{+}\text{ particles released = }\left(\text{4}\times \text{Atomic no}\text{.of }H\right)-\left(\text{Atomic no}\text{. of }He\right)\\ \\ \text{= }\left[\left(4\times 1\right)\right]-2\\ \\ =\text{ 2}\end{array}$

Atomic number of each positron particle is 1. Hence, number of positron particle is 2.

Therefore, equation can be given as,

  $4{}_1^{1}H\underset{}{\to }{}_2^{4}He{\text{ + 2 }}_1^0\beta$

(d)

Interpretation Introduction

Interpretation:

Changes in mass per kilogram of dilithium and of Helium-4 have to be compared.

Concept Introduction:

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  $\text{Nucleus + nuclear binding energy }\underset{}{\to }\text{ nucleons}$

It can be calculated using the given formula,

  $\begin{array}{l}\Delta \text{E = }\Delta {\text{mc}}^2\\ \\ where,\\ \Delta \text{m = Mass Difference}\\ c=\text{ Speed of light}\end{array}$

Change in mass of a given reaction can be determined as given,

  $\Delta m\text{ = }Mass\text{ of reactant}-\text{ Mass of products}$

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}Massdifference{\text{ for }}_6^{12}\text{C}=\text{ 5}\text{.02180507}\times {10}^{-29}\text{ kg/atom}\\ \\ {\text{Mass of }}_1^{1}H=\text{ 1}\text{.007825 amu}\\ \\ {\text{Mass of }}_2^{4}He=\text{ 4}\text{.00260 amu}\\ \\ Mass\text{ of }{\beta }^{+}\text{ = Mass of an electron = 5}\text{.48580}\times {\text{10}}^{-4}\text{ amu}\end{array}$

• Calculate change in mass per kilogram of dilithium:

Change in mass per kilogram of dilithium is determined as follows,

  $\begin{array}{c}\Delta m\text{ = }\left(\frac{\text{5}\text{.02180507}\times {10}^{-29}\text{ kg}}{atom}\right)\left(\frac{1\text{ atom}}{\text{12}\text{.000000 amu}}\right)\left(\frac{1\text{ amu}}{1.66054\times {10}^{-27}{\text{ kg }}^{12}C}\right)\\ \\ =\text{ 2}\text{.5202}\times {\text{10}}^{-3}{\text{ kg/kg }}^{12}C\end{array}$

Mass per kilogram of dilithium is $2.5202\times 10^{-3}kg/kg{}^{12}C.$

• Calculate the mass difference for the formation of Helium-4:

Mass difference of the reaction can be calculated as given,

$\begin{array}{c}\Delta m\text{ = }Mass\text{ of reactant}-\text{ Mass of products}\\ \\ \text{= }\left[Mass{\text{ of }}_1^{1}H\right]-\left[{\text{Mass of }}_2^4\text{He + }\left(\text{2}\times \text{Mass of }{\beta }^{+}\right)\right]\\ \\ \text{= }\left[\text{1}\text{.007825 amu}\right]-\left[\left(\text{4}\text{.00260 amu}\right)+\left(2\times \text{5}\text{.48580}\times {\text{10}}^{-4}\text{ amu}\right)\right]\\ \\ =\text{0}\text{.02760 amu/atom}=\left(\text{0}\text{.02760 amu/atom}\right)\left(1.66054\times {10}^{-27}\text{ kg/amu}\right)\\ \\ =\text{ 4}\text{.58309}\times {10}^{-29}\text{ kg/atom}\end{array}$

Mass difference for the formation of Helium-4 is $\text{ 3}\text{.1218152}\times {10}^{-29}\text{ kg/atom}$

• Calculate change in mass per kilogram of Helium-4:

Change in mass per kilogram of Helium-4 is determined as follows,

  $\begin{array}{c}\Delta m\text{ = }\left(\frac{\text{4}\text{.58309}\times {10}^{-29}\text{ kg}}{atom}\right)\left(\frac{1\text{ atom}}{\text{12}\text{.000000 amu}}\right)\left(\frac{1\text{ amu}}{1.66054\times {10}^{-27}{\text{ kg }}^{12}C}\right)\\ \\ =\text{ 6}\text{.896}\times {\text{10}}^{-3}{\text{ kg/kg }}^{4}He\end{array}$

Mass per kilogram of Helium-4 is $6.896\times 10^{-3}kg/kg{}^{4}He.$

Comparing both the values, mass per kilogram of Helium-4 is much higher than mass per kilogram of dilithium.

(e)

Interpretation Introduction

Interpretation:

Change in mass per kilogram for method used in current fusion reactors has to be compared.

Concept Introduction:

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  $\text{Nucleus + nuclear binding energy }\underset{}{\to }\text{ nucleons}$

It can be calculated using the given formula,

  $\begin{array}{l}\Delta \text{E = }\Delta {\text{mc}}^2\\ \\ where,\\ \Delta \text{m = Mass Difference}\\ c=\text{ Speed of light}\end{array}$

Change in mass of a given reaction can be determined as given,

  $\Delta m\text{ = }Mass\text{ of reactant}-\text{ Mass of products}$

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}Given{\text{ reaction: }}_1^{3}H{\text{ + }}_1^{2}H\underset{}{\to }{}_2^{4}He+{}_0^{1}n\\ \\ {\text{Mass of }}_1^{3}H=\text{ 3}\text{.01605 amu}\\ \\ {\text{Mass of }}_1^{2}H=\text{ 2}\text{.0140 amu}\\ \\ {\text{Mass of }}_2^{4}He=\text{ 4}\text{.0026 amu}\\ \\ {\text{Mass of }}_0^{1}n\text{ = 1}\text{.008665 amu}\end{array}$

• Calculate the mass of reactants and products:

Mass of reactants is determined as shown below,

  $\begin{array}{c}Mass{\text{ of reactants = Mass of }}_1^{3}H{\text{ + Mass of }}_1^{2}H\\ \\ =\left(\text{3}\text{.01605 amu}\right)+\left(\text{2}\text{.0140 amu}\right)\\ \\ =\text{ 5}\text{.03005 amu}\end{array}$

Mass of products is determined as shown below,

  $\begin{array}{c}Mass{\text{ of products = Mass of }}_2^{4}He{\text{ + Mass of }}_0^{1}n\\ \\ =\left(\text{4}\text{.0026 amu}\right)+\left(\text{1}\text{.008665 amu}\right)\\ \\ =\text{ 5}\text{.011265 amu}\end{array}$

• Calculate the mass difference:

Mass difference of the reaction can be calculated as given,

  $\begin{array}{c}\Delta m\text{ = }Mass\text{ of reactant}-\text{ Mass of products}\\ \\ \text{= }\left(\text{5}\text{.03005 amu}\right)-\left(\text{5}\text{.011265 amu}\right)\\ \\ =\text{ 0}\text{.0188 amu}\\ \\ \text{ = }\left(\text{0}\text{.0188 amu/atom}\right)\left(1.66054\times {10}^{-27}\text{ kg/amu}\right)\\ \\ =\text{ 3}\text{.1218152}\times {10}^{-29}\text{ kg/atom}\end{array}$

Mass difference is $\text{3}\text{.1218152}\times {10}^{-29}\text{ kg/atom}$

• Calculate change in mass per kilogram of Helium-4 used in current fusion reactors:

Change in mass per kilogram is determined as follows,

  $\begin{array}{c}\Delta m\text{ = }\left(\frac{\text{3}\text{.1218152}\times {10}^{-29}\text{ kg}}{atom}\right)\left(\frac{1\text{ atom}}{\text{12}\text{.000000 amu}}\right)\left(\frac{1\text{ amu}}{1.66054\times {10}^{-27}{\text{ kg }}^{12}C}\right)\\ \\ =\text{ 4}\text{.696947}\times {\text{10}}^{-3}{\text{ kg/kg }}^{4}He\end{array}$

Change in mass per kilogram of Helium-4 used in current fusion reactors is $\text{4}\text{.696947}\times {\text{10}}^{-3}{\text{ kg/kg }}^{4}He.$

Mass per kilogram of dilithium is $2.5202\times 10^{-3}kg/kg{}^{12}C.$

Comparing both the values, mass per kilogram of Helium-4 used in current fusion reactors is much higher than mass per kilogram of dilithium.

(f)

Interpretation Introduction

Interpretation:

Change in mass of the given reaction has to be determined and compared with the value of dilithium reaction.

Concept Introduction:

Nuclear binding energy: It is the energy that requires for the breaking one mole of nuclei of an element to its individual nucleons.

  $\text{Nucleus + nuclear binding energy }\underset{}{\to }\text{ nucleons}$

It can be calculated using the given formula,

  $\begin{array}{l}\Delta \text{E = }\Delta {\text{mc}}^2\\ \\ where,\\ \Delta \text{m = Mass Difference}\\ c=\text{ Speed of light}\end{array}$

Change in mass of a given reaction can be determined as given,

  $\Delta m\text{ = }Mass\text{ of reactant}-\text{ Mass of products}$

Explanation of Solution

Given reaction is ${}_3^{6}Li{\text{ + }}_0^{1}n\underset{}{\to }{}_2^{4}He{\text{ + }}_1^{3}H$

Mass per kilogram of ${}_1^{3}H$ formed is calculated as follows,

$\begin{array}{c}Mass{}_1^{3}H/kg{}^{6}Li\text{ = }\left(\frac{1{\text{ mol }}^{3}H}{1{\text{ mol }}^{6}Li}\right)\left(\frac{3.01605{\text{ g }}^{3}H}{1{\text{ mol }}^{3}H}\right)\left(\frac{1{\text{ mol }}^{6}Li}{\text{6}{\text{.015121 g }}^6\text{Li}}\right)\left(\frac{{10}^3{\text{ g }}^{6}Li}{1{\text{ kg }}^{6}Li}\right)\left(\frac{1{\text{ kg }}^{3}H}{{10}^3{\text{ g }}^{3}H}\right)\\ \\ =\text{ 0}{\text{.501411 kg }}^{\text{3}}{\text{H/kg }}^{\text{6}}\text{Li}\end{array}$

Mass per kilogram of ${}_1^{3}H$ formed is $\text{0}{\text{.501411 kg }}^{\text{3}}{\text{H/kg }}^{\text{6}}\text{Li}\text{.}$

Mass of ${}_1^{3}H$ formed is calculated as follows,

$\begin{array}{c}Mass\text{ = }\left(\text{0}{\text{.501411 kg }}^{\text{3}}\text{H}\right)\left(\frac{{10}^3\text{ g}}{\text{1 kg}}\right)\left(\frac{{\text{1 mol }}^3\text{H}}{\text{3}{\text{.01605 g }}^3\text{H}}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}{\text{ atom }}^3\text{H}}{{\text{mol }}^3\text{H}}\right)\left(\frac{\text{3}\text{.121852}\times {10}^{-29}\text{ kg}}{atom}\right)\\ \\ =\text{ 3}\text{.125}\times {\text{10}}^{-3}\text{ kg}\end{array}$

Mass of ${}_1^{3}H$ formed is $\text{3}\text{.125}\times {\text{10}}^{-3}\text{ kg}$

Change in mass for dilithium reaction is shown below,

  $\begin{array}{c}Mass\text{ = }\left[\begin{array}{l}\left(\frac{\text{5}\text{.02180507}\times {10}^{-29}{\text{ kg }}^{12}C}{atom}\right)\left(\frac{1atom{}^{12}C}{{\text{2 atoms }}^{6}Li}\right)\left(\frac{6.022\times {10}^{23}{\text{ atoms }}^{6}Li}{mol}\right)\\ \left(\frac{1{\text{ mol }}^{6}Li}{\text{6}{\text{.015121 g }}^{\text{6}}\text{Li}}\right)\left(\frac{{10}^3{\text{ g }}^{6}Li}{1{\text{ kg }}^{6}Li}\right)\end{array}\right]\\ \\ =\text{ 2}\text{.514}\times {\text{10}}^{-3}\text{ kg}\end{array}$

Change in mass for dilithium reaction is $\text{2}\text{.514}\times {\text{10}}^{-3}\text{ kg}$

Comparing to change in mass for the dilithium reaction, change in mass for the fusion of tritium with deuterium is slightly high.