Chapter 24, Problem 24.127P

(a)

Interpretation Introduction

Interpretation:

Kinetic Energy of an $H$ atom has to be determined using the given data.

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.

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

Kinetic energy of a particle can be using the below mentioned formula,

  $KE=\frac{1}{2}m{v}^2=\left(\frac{3}{2}\right)\frac{RT}{N_A}\boxed{\begin{array}{l}R=\text{ Gas constant}\\ \text{T = Temperature}\\ {\text{N}}_A\text{ = Avogadro's number}\end{array}}$

Explanation of Solution

Given data is shown below:

$Temperature,\text{ T = 1}\text{.00}\times {\text{10}}^6\text{ K}$

Kinetic Energy of an $H$ atom is calculated as shown below,

  $\begin{array}{c}Kinetic\text{ Energy = }\left(\frac{3}{2}\right)\frac{RT}{N_A}\\ \\ \text{= }\left(\frac{3}{2}\right)\frac{\left(8.314\text{ J/mol}\text{.K}\right)\left(1.00\times {10}^{6}K\right)}{\left(6.022\times {10}^{23}\text{ atom/mol}\right)}\\ \\ =\text{ 2}\text{.07}\times {\text{10}}^{-17}{\text{ J/atom }}_1^{1}H\end{array}$

Kinetic Energy of an $H$ atom is $2.07\times 10^{-17}J/atom.$

(b)

Interpretation Introduction

Interpretation:

Number of $H$ atoms heated from the energy produced by the annihilation of one $H$ and one $antiH$ has to be determined using the given data.

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.

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

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,

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

Explanation of Solution

Given data is shown below:

Kinetic Energy of an $H$ atom is $\text{2}{\text{.07×10}}^{\text{-17}}\text{ J/atom}$

A kilogram of $antiH$ gets annihilated by a kilogram of ${}^{1}H$. Hence $2\text{ kg}$ is converted to energy and it can be determined as follows,

  $\begin{array}{c}Energy{\text{ = mc}}^2\\ \\ =\left(2.00\text{ kg}\right){\left(2.998\times {10}^8{\text{ ms}}^{-1}\right)}^2\left(J/kg.m^2{s}^{-2}\right)\\ \\ =\text{ 1}\text{.7975048}\times {10}^{17}\text{ J per one kg H}\end{array}$

Energy produced by the annihilation of one $H$ and one $antiH$ is $\text{1}\text{.7975048}\times {10}^{17}\text{ J}$

Number of $H$ atoms heated from the energy produced by the annihilation of one $H$ and one $antiH$ is calculated as shown below,

  $\begin{array}{c}No.\text{ of atoms = }\left[\begin{array}{l}\left(\frac{\text{1}{\text{.7975048×10}}^{\text{17}}\text{ J}}{1\text{ kg H}}\right)\left(\frac{1\text{ kg}}{{10}^3\text{ g}}\right)\left(\frac{\text{1}\text{.0078 g H}}{\text{1 mol H}}\right)\\ \left(\frac{\text{1 mol H}}{6.022\times {10}^{23}\text{ atoms H}}\right)\left(\frac{\text{1 H atom}}{\text{2}{\text{.07×10}}^{-\text{17}}\text{ J/atom}}\right)\end{array}\right]\\ \\ =\text{ 1}\text{.45}\times {\text{10}}^7\text{ H atoms}\end{array}$

Number of $H$ atoms is $1.45\times 10^{7}Hatoms.$

(c)

Interpretation Introduction

Interpretation:

Energy released during the fusion of 4 Hydrogen atoms forming Helium-4 has to be determined using the given data.

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.

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

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,

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

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 data is shown below:

  $\begin{array}{c}\text{Reaction: 4}{}_{\text{1}}^{\text{1}}\text{H }\underset{}{\to }{}_{\text{2}}^{\text{4}}\text{He + 2}{}_{\text{+1}}^{\text{0}}\text{β}\\ \\ \text{Number of H atoms = 1}\text{.45}\times {\text{10}}^7\\ \\ {\text{Mass of }}_1^{1}H=\text{ 1}\text{.007825 amu}\\ \\ {\text{Mass of }}_2^{4}He=\text{ 4}\text{.00260 amu}\\ \\ Mass\text{ of }{\beta }^{+}\text{ = Mass of an electron = 5}\text{.48580}\times {\text{10}}^{-4}\text{ amu}\end{array}$

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

Mass difference of the reaction can be calculated as given,

  $\begin{array}{c}\Delta m\text{ = }Mass\text{ of reactant}-\text{ Mass of products}\\ \\ \text{= }\left[4\times Mass{\text{ of }}_1^{1}H\right]-\left[{\text{Mass of }}_2^4\text{He + }\left(\text{2}\times \text{Mass of }{\beta }^{+}\right)\right]\\ \\ \text{= }\left[\text{4}\times \text{1}\text{.007825 amu}\right]-\left[\left(\text{4}\text{.00260 amu}\right)+\left(2\times \text{5}\text{.48580}\times {\text{10}}^{-4}\text{ amu}\right)\right]\\ \\ =\text{0}{\text{.02760 amu per }}_2^{4}He\text{ formed}\\ \\ \text{= 0}{\text{.02760 g per mol }}_2^{4}He\text{ formed}\end{array}$

Mass difference during the fusion of 4 Hydrogen atoms is $\text{0}\text{.02760 g per mol}\text{.}$

• Calculate total mass difference in kilogram:

Mass difference during the fusion of $\text{1}\text{.45}\times {\text{10}}^7$ atoms is calculated as follows,

$\begin{array}{c}\Delta \text{m = }\left(\frac{\text{0}\text{.02760 amu}}{mo{l}^{4}He}\right)\left(\frac{1\text{ kg}}{{10}^3\text{ g}}\right)\left(\frac{1{}^{4}He}{4\text{ H atoms}}\right)\left(\frac{1\text{ mol H}}{6.022\times {10}^{23}\text{ H atoms}}\right)\left(\text{1}{\text{.45×10}}^{\text{7}}\text{ H atoms}\right)\\ \\ =\text{ 1}\text{.6615}\times {\text{10}}^{-22}\text{ kg}\end{array}$

• Calculate energy released:

Energy released is calculated as given below,

  $\begin{array}{c}Energy\text{ = }\Delta m{c}^2\\ \\ \text{= }\left(\text{1}\text{.6615}\times {\text{10}}^{-22}\text{ kg}\right)\left(2.998\times {10}^8{\text{ ms}}^{-1}\right)\\ \\ =\text{ 1}\text{.49}\times {\text{10}}^{-5}\text{ J}\end{array}$

Energy released during the fusion of 4 Hydrogen atoms forming Helium-4 is $1.49\times 10^{-5}J.$

(d)

Interpretation Introduction

Interpretation:

Increase in energy produce by the fusion of part (c) comparing with $H-antiH$ collision of part (b) has to be determined using the given data.

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.

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

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,

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

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

In part (b), Energy produced by the annihilation of one $H$ and one $antiH$ is $\text{1}\text{.7975048}\times {10}^{17}\text{ J }$

Therefore, energy generated by $H-antiH$ collision is determined as below,

  $\begin{array}{c}Energy\text{ = }\left(\frac{\text{1}\text{.7975048}\times {10}^{17}\text{ J }}{1\text{ kg H}}\right)\left(\frac{1\text{ kg}}{{10}^3\text{ g}}\right)\left(\frac{\text{1}\text{.0078 g H}}{\text{1 mol H}}\right)\left(\frac{\text{1 mol H}}{6.022\times {10}^{23}\text{ atoms H}}\right)\\ \\ \text{ = 3}\text{.00}\times {\text{10}}^{-10}\text{ J}\end{array}$

In part (c), Energy released during the fusion of 4 Hydrogen atoms forming Helium-4 is $\text{1}{\text{.49×10}}^{\text{-5}}\text{ J}\text{.}$

Increase in energy is determined as follows,

$\begin{array}{c}Energy\text{ increase = }\left(\text{1}\text{.49}\times {\text{10}}^{-5}\text{ J}\right)-\left(\text{3}\text{.00}\times {\text{10}}^{-10}\text{ J}\right)\\ \\ =\text{ 1}\text{.49}\times {\text{10}}^{-5}\text{ J}\end{array}$

Energy increase is $1.49\times 10^{-5}J.$

(e)

Interpretation Introduction

Interpretation:

Energy released when $1.00kg$ each of antimatter and matter annihilate with each other has to be determined using the given data.

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.

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

Nuclear binding energy can be calculated by Einstein’s mass energy equivalence relationship that is,

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

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 data is shown below:

  $\begin{array}{c}\text{Reaction: 3}{}_{\text{1}}^{\text{1}}\text{H }\underset{}{\to }{}_{\text{2}}^3\text{He + 1}{}_{\text{+1}}^{\text{0}}\text{β}\\ \\ \text{Number of H atoms = 1}\text{.45}\times {\text{10}}^7\\ \\ {\text{Mass of }}_1^{1}H=\text{ 1}\text{.007825 amu}\\ \\ {\text{Mass of }}_2^{3}He=\text{ 3}\text{.01603 amu}\\ \\ Mass\text{ of }{\beta }^{+}\text{ = Mass of an electron = 5}\text{.48580}\times {\text{10}}^{-4}\text{ amu}\end{array}$

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

Mass difference of the reaction can be calculated as given,

  $\begin{array}{c}\Delta m\text{ = }Mass\text{ of reactant}-\text{ Mass of products}\\ \\ \text{= }\left[3\times Mass{\text{ of }}_1^{1}H\right]-\left[{\text{Mass of }}_2^3\text{He + Mass of }{\beta }^{+}\right]\\ \\ \text{= }\left[\text{3}\times \text{1}\text{.007825 amu}\right]-\left[\text{3}\text{.01603}+\left(\text{5}\text{.48580}\times {\text{10}}^{-4}\text{ amu}\right)\right]\\ \\ =\text{0}{\text{.006896 amu per }}_2^3\text{He formed}\\ \\ \text{= 0}{\text{.006896 g/mol }}_2^3\text{He formed}\end{array}$

Mass difference during the fusion of 3 Hydrogen atoms is $\text{0}\text{.006896 g/mol}\text{.}$

• Calculate total mass difference in kilogram:

Mass difference during the fusion of $\text{1}\text{.45}\times {\text{10}}^7$ atoms is calculated as follows,

$\begin{array}{c}\Delta \text{m = }\left(\frac{\text{0}\text{.006896 amu}}{mo{l}^{4}He}\right)\left(\frac{1\text{ kg}}{{10}^3\text{ g}}\right)\left(\frac{1{}^{3}He}{\text{3 H atoms}}\right)\left(\frac{1\text{ mol H}}{6.022\times {10}^{23}\text{ H atoms}}\right)\left(\text{1}{\text{.45×10}}^{\text{7}}\text{ H atoms}\right)\\ \\ =\text{ 5}\text{.5348}\times {\text{10}}^{-23}\text{ kg}\end{array}$

• Calculate energy released:

Energy released is calculated as given below,

  $\begin{array}{c}Energy\text{ = }\Delta m{c}^2\\ \\ \text{= }\left(\text{5}\text{.5348}\times {\text{10}}^{-23}\text{ kg}\right)\left(2.998\times {10}^8{\text{ ms}}^{-1}\right)\\ \\ =\text{ 4}\text{.97}\times {\text{10}}^{-6}\text{ J}\end{array}$

Energy released during the fusion of 3 Hydrogen atoms forming Helium-3 is $4.97\times 10^{-6}J.$

Energy released during the fusion of 4 Hydrogen atoms forming Helium-4 is $1.49\times 10^{-5}J.$ It is higher comparing to the energy produced during the formation of Helium-3.

Therefore, Chief Engineer should advise Captain not to change the technology and to keep the current technology.