CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
8th Edition
ISBN: 9781309097182
Author: SILBERBERG
Publisher: MCG/CREATE
Question
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Chapter 24, Problem 24.114P

(a)

Interpretation Introduction

Interpretation:

Product formed by the fusion of two 6Li 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.

(a)

Expert Solution
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Explanation of Solution

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

Therefore, the reaction can be given as,

  36Li + 36Li  612C(dilithium)

(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.

  Nucleus + nuclear binding energy  nucleons

It can be calculated using the given formula,

  ΔE = Δmc2where,Δm = Mass Differencec= Speed of light

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

  Δm = Mass of reactant Mass of products

(b)

Expert Solution
Check Mark

Explanation of Solution

Given information is shown below,

  Given reaction: 36Li + 36Li  612CMass  of 36Li= 6.015121 amuMass  of 612C= 12.000000 amu

  • Calculate the mass difference:

Mass difference of the reaction can be calculated as given,

  Δm = Mass of reactant Mass of products[2×Mass of 36Li] [ Mass of 612C][2×6.015121 amu][12.000000 amu]= 0.030242 amu/atom  612C

Mass difference is 0.030242 amu/atom.

  • Convert the unit of mass difference:

Unit of mass difference is converted from amu/atom to kg/atom as follows,

  Δm = (0.030242 amu/atom)(1.66054×1027 kg/amu)= 5.02180507×1029 kg/atom

  • Calculate the energy per atom:

Energy per atom is calculated as follows for the reaction,

Binding energy, E = Δmc2= (5.02180507×1029 kg/atom)(2.998×108 m/s)2(1 J1 kg.m2.s2)= 4.5133595×1012 J/atom

Energy per atom of the given reaction is 4.5133595×1012 J/atom.

  • Convert the unit of energy:

Unit of energy is converted from J/atom to J/kg as given below,

  E = (4.5133595×1012 Jatom)(1 atom12.000000 amu)(1 amu1.66054×1027 kg)= 2.2650×1014 J/kg dilithium

Therefore,

Energy released during the reaction is 2.2650×1014 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,

  ParticleSymbolProton11Hor11PNeutron01nElectron-10eAlphaparticle24Heor24αBetaparticle-10eor-10βPositron10e

(c)

Expert Solution
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Explanation of Solution

The given unbalanced nuclear equation is,

  4 11H  24He + x 10β

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.

Total atomic no. for β+ particles released = (4×Atomic no.of H)(Atomic no. of He)[(4×1)]2= 2

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

Therefore, equation can be given as,

  4 11H  24He + 2 10β

(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.

  Nucleus + nuclear binding energy  nucleons

It can be calculated using the given formula,

  ΔE = Δmc2where,Δm = Mass Differencec= Speed of light

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

  Δm = Mass of reactant Mass of products

(d)

Expert Solution
Check Mark

Explanation of Solution

Given information is shown below,

  Mass difference for 612C= 5.02180507×1029 kg/atomMass  of 11H= 1.007825 amuMass  of 24He= 4.00260 amuMass of β+ = Mass of an electron = 5.48580×104 amu

  • Calculate change in mass per kilogram of dilithium:

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

  Δm = (5.02180507×1029 kgatom)(1 atom12.000000 amu)(1 amu1.66054×1027 kg 12C)= 2.5202×103 kg/kg 12C

Mass per kilogram of dilithium is 2.5202×10-3 kg/kg 12C.

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

Mass difference of the reaction can be calculated as given,

Δm = Mass of reactant Mass of products[Mass of 11H] [Mass of 24He + (2×Mass of β+)][1.007825 amu][(4.00260 amu)+(2×5.48580×104 amu)]=0.02760 amu/atom= (0.02760 amu/atom)(1.66054×1027 kg/amu)= 4.58309×1029 kg/atom

Mass difference for the formation of Helium-4 is  3.1218152×1029 kg/atom

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

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

  Δm = (4.58309×1029 kgatom)(1 atom12.000000 amu)(1 amu1.66054×1027 kg 12C)= 6.896×103 kg/kg 4He

Mass per kilogram of Helium-4 is 6.896×10-3 kg/kg 4He.

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.

  Nucleus + nuclear binding energy  nucleons

It can be calculated using the given formula,

  ΔE = Δmc2where,Δm = Mass Differencec= Speed of light

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

  Δm = Mass of reactant Mass of products

(e)

Expert Solution
Check Mark

Explanation of Solution

Given information is shown below,

  Given reaction: 13H + 12H  24He + 01n Mass  of 13H= 3.01605 amuMass  of 12H= 2.0140 amuMass  of 24He= 4.0026 amuMass  of 01n = 1.008665 amu

  • Calculate the mass of reactants and products:

Mass of reactants is determined as shown below,

  Mass of reactants = Mass of 13H + Mass  of 12H= (3.01605 amu)+(2.0140 amu)= 5.03005 amu

Mass of products is determined as shown below,

  Mass of products = Mass  of 24He + Mass  of 01n= (4.0026 amu)+(1.008665 amu)= 5.011265 amu

  • Calculate the mass difference:

Mass difference of the reaction can be calculated as given,

  Δm = Mass of reactant Mass of products(5.03005 amu)(5.011265 amu)= 0.0188 amu = (0.0188 amu/atom)(1.66054×1027 kg/amu)= 3.1218152×1029 kg/atom

Mass difference is 3.1218152×1029 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,

  Δm = (3.1218152×1029 kgatom)(1 atom12.000000 amu)(1 amu1.66054×1027 kg 12C)= 4.696947×103 kg/kg 4He

Change in mass per kilogram of Helium-4 used in current fusion reactors is 4.696947×103 kg/kg 4He.

Mass per kilogram of dilithium is 2.5202×10-3 kg/kg 12C.

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.

  Nucleus + nuclear binding energy  nucleons

It can be calculated using the given formula,

  ΔE = Δmc2where,Δm = Mass Differencec= Speed of light

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

  Δm = Mass of reactant Mass of products

(f)

Expert Solution
Check Mark

Explanation of Solution

Given reaction is 36Li + 01n  24He + 13H

Mass per kilogram of 13H formed is calculated as follows,

Mass 13H/kg 6Li = (1 mol 3H1 mol 6Li)(3.01605 g 3H1 mol 3H)(1 mol 6Li6.015121 g 6Li)(103 g 6Li1 kg 6Li)(1 kg 3H103 g 3H)= 0.501411 kg 3H/kg 6Li

Mass per kilogram of 13H formed is 0.501411 kg 3H/kg 6Li.

Mass of 13H formed is calculated as follows,

Mass = (0.501411 kg 3H)(103 g1 kg)(1 mol 3H3.01605 g 3H)(6.022×1023 atom 3Hmol 3H)(3.121852×1029 kgatom)= 3.125×103 kg

Mass of 13H formed is 3.125×103 kg

Change in mass for dilithium reaction is shown below,

  Mass = [(5.02180507×1029 kg 12Catom)(1 atom 12C2 atoms 6Li)(6.022×1023 atoms 6Limol)(1 mol 6Li6.015121 g 6Li)(103 g 6Li1 kg 6Li)]= 2.514×103 kg

Change in mass for dilithium reaction is 2.514×103 kg

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

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Chapter 24 Solutions

CHEMISTRY >CUSTOM<

Ch. 24.2 - Prob. 24.6AFPCh. 24.2 - Prob. 24.6BFPCh. 24.6 - Prob. 24.7AFPCh. 24.6 - Prob. 24.7BFPCh. 24.7 - Prob. B24.1PCh. 24.7 - Prob. B24.2PCh. 24.7 - Prob. B24.3PCh. 24.7 - Prob. B24.4PCh. 24 - Prob. 24.1PCh. 24 - Prob. 24.2PCh. 24 - Prob. 24.3PCh. 24 - Prob. 24.4PCh. 24 - Prob. 24.5PCh. 24 - Prob. 24.6PCh. 24 - Prob. 24.7PCh. 24 - Prob. 24.8PCh. 24 - Prob. 24.9PCh. 24 - Prob. 24.10PCh. 24 - Prob. 24.11PCh. 24 - Prob. 24.12PCh. 24 - Prob. 24.13PCh. 24 - Prob. 24.14PCh. 24 - Prob. 24.15PCh. 24 - Prob. 24.16PCh. 24 - Prob. 24.17PCh. 24 - Prob. 24.18PCh. 24 - Prob. 24.19PCh. 24 - Prob. 24.20PCh. 24 - Prob. 24.21PCh. 24 - Prob. 24.22PCh. 24 - Prob. 24.23PCh. 24 - Prob. 24.24PCh. 24 - Prob. 24.25PCh. 24 - Prob. 24.26PCh. 24 - Prob. 24.27PCh. 24 - Prob. 24.28PCh. 24 - Prob. 24.29PCh. 24 - Prob. 24.30PCh. 24 - Prob. 24.31PCh. 24 - Prob. 24.32PCh. 24 - Prob. 24.33PCh. 24 - Prob. 24.34PCh. 24 - Prob. 24.35PCh. 24 - Prob. 24.36PCh. 24 - Prob. 24.37PCh. 24 - Prob. 24.38PCh. 24 - Prob. 24.39PCh. 24 - Prob. 24.40PCh. 24 - Prob. 24.41PCh. 24 - Prob. 24.42PCh. 24 - Prob. 24.43PCh. 24 - Prob. 24.44PCh. 24 - Prob. 24.45PCh. 24 - Prob. 24.46PCh. 24 - Prob. 24.47PCh. 24 - Prob. 24.48PCh. 24 - Prob. 24.49PCh. 24 - Prob. 24.50PCh. 24 - Prob. 24.51PCh. 24 - Prob. 24.52PCh. 24 - Prob. 24.53PCh. 24 - Prob. 24.54PCh. 24 - Prob. 24.55PCh. 24 - Prob. 24.56PCh. 24 - Prob. 24.57PCh. 24 - Prob. 24.58PCh. 24 - Prob. 24.59PCh. 24 - Prob. 24.60PCh. 24 - Prob. 24.61PCh. 24 - Prob. 24.62PCh. 24 - Prob. 24.63PCh. 24 - Prob. 24.64PCh. 24 - Prob. 24.65PCh. 24 - Prob. 24.66PCh. 24 - Prob. 24.67PCh. 24 - Prob. 24.68PCh. 24 - Prob. 24.69PCh. 24 - Prob. 24.70PCh. 24 - Prob. 24.71PCh. 24 - Prob. 24.72PCh. 24 - Prob. 24.73PCh. 24 - Prob. 24.74PCh. 24 - Prob. 24.75PCh. 24 - Prob. 24.76PCh. 24 - Prob. 24.77PCh. 24 - Prob. 24.78PCh. 24 - Prob. 24.79PCh. 24 - Prob. 24.80PCh. 24 - Prob. 24.81PCh. 24 - Prob. 24.82PCh. 24 - Prob. 24.83PCh. 24 - Prob. 24.84PCh. 24 - Prob. 24.85PCh. 24 - Prob. 24.86PCh. 24 - Prob. 24.87PCh. 24 - Prob. 24.88PCh. 24 - Prob. 24.89PCh. 24 - Prob. 24.90PCh. 24 - Prob. 24.91PCh. 24 - Prob. 24.92PCh. 24 - Prob. 24.93PCh. 24 - Prob. 24.94PCh. 24 - Prob. 24.95PCh. 24 - Prob. 24.96PCh. 24 - Prob. 24.97PCh. 24 - Prob. 24.98PCh. 24 - Prob. 24.99PCh. 24 - Prob. 24.100PCh. 24 - Prob. 24.101PCh. 24 - Prob. 24.102PCh. 24 - Prob. 24.103PCh. 24 - Prob. 24.104PCh. 24 - Prob. 24.105PCh. 24 - Prob. 24.106PCh. 24 - Prob. 24.107PCh. 24 - Prob. 24.108PCh. 24 - Prob. 24.109PCh. 24 - Prob. 24.110PCh. 24 - Prob. 24.111PCh. 24 - Prob. 24.112PCh. 24 - Prob. 24.113PCh. 24 - Prob. 24.114PCh. 24 - Prob. 24.115PCh. 24 - Prob. 24.116PCh. 24 - Prob. 24.117PCh. 24 - Prob. 24.118PCh. 24 - Prob. 24.119PCh. 24 - Prob. 24.120PCh. 24 - Prob. 24.121PCh. 24 - Prob. 24.122PCh. 24 - Prob. 24.123PCh. 24 - Prob. 24.124PCh. 24 - Prob. 24.125PCh. 24 - Prob. 24.126PCh. 24 - Prob. 24.127PCh. 24 - Prob. 24.128PCh. 24 - Prob. 24.129PCh. 24 - Prob. 24.130PCh. 24 - Prob. 24.131PCh. 24 - Prob. 24.132PCh. 24 - Prob. 24.133PCh. 24 - Prob. 24.134PCh. 24 - Prob. 24.135PCh. 24 - Prob. 24.136PCh. 24 - Prob. 24.137PCh. 24 - Prob. 24.138PCh. 24 - Prob. 24.139PCh. 24 - Prob. 24.140PCh. 24 - Prob. 24.141PCh. 24 - Prob. 24.142PCh. 24 - Prob. 24.143PCh. 24 - Prob. 24.144P
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