Homework(Solutions(:(Chapter(32((1.##Picture(the(Problem:#We#are#given#three#isotope#designations#and#asked#to#calculate#the#atomic#number#(A),#neutron#number#(N),#and#atomic#mass#(Z)#for#each#isotope .##Strategy:#Use#the#designation#of#an#isotope#is#,AZX#where#A#is#the #at omic#mass,#Z#is#the#atom ic#n u mber,#and#X%is#the#chem ic al#d e sig n at io n ,#to#d et er m in e #th e#a to mic#numbe r#a n d #ato mic#mass#of#th e#th re e #iso to p es .#Ca lcu l at e#the#neutron#num be r#by#sub tracting #the#atomic#number#from#the#ato m ic#m ass.###Solution:#1.((a)(Write#Z,%N,#and#A#for#the #isot op e #23892U:#92Z = 238 92 146NAZ=−= − = 238A =##2.#(b)(Write#Z,%N,#and#A#for#the#isotop e#239 94Pu:#94Z = 239 94 145NAZ=−= − = 239A =##3.((c)#Write#Z,%N,#and#A%for#the#isotope #144 60Nd :#60Z =#144 60 84NAZ=−= − =#144A =##Insight:#Note#that#for#each#of#these#isotopes#the#neutron#number#is#about#1½#times#the#atomic#number.#The#neutron#number#is#about#equal#to#the#atomic#number#for#small#isotopes#and#about#1½#times#the#atomic#number#for#large#isotopes.#(4.##Picture(the(Problem:#The#number#of#neutrons#in#a#chlorine#nu c le us #ca n #b e#d et er m in e d #fro m #its #n u cle a r#radius.##Strategy: Use equation 32-4 to calculate the mass number of the chlorine atom. The periodic table shows that the atomic number of chlorine is 17. Subtract the atomic number from the atomic mass to calculate the neutron number. #Solution:#1.(Solve#equation#32H4#for#the#atomic#mass:(#( )3315315154.0 10 m371.2 10 m1.2 10 mrA−−−⎛⎞×== =⎜⎟×⎝⎠×##2.#Solve#equation#32H1#for#the#neutron#number:# 37 17 20AZN N AZ=+ ⇒ =−= − =##Insight:#At#best#we#can#say#there#are#between#19#and#21#neutrons#because#the#radius#is#only#given#to#two#significant#figures.#For#instance,#note#that#adding#one#additional#neutron#to#make #A#=#38#also#gives#a#radius#of#154.0 10 m.−×###5.##Picture(the(Problem:#The#nuclear#densities#of#thoriumH238#and#an#alpha#particle#(heliumH4)#can#be#compared.##Strategy: The nuclear density is the mass of the nucleus divided by its volume. Calculate the mass by multiplying the mass number by the mass of one nucleon. The volume is the volume of a sphere whose radius is given by equation 33-4. #Solution:#1.((a)#Write#the#density#as#mass#over#volume:##343MAmVrρπ==##2.(Use#equation#33H4#to#write#the#radius#and#simplify:((( )1333 3000333444Am Am mrA rrAρπππ===##3.(Insert#the#constants#to#calcula te##the#nuclear#density#of(22890Th:(( )( )2717 33153 1.67 10 kg2.3 10 kg/m4 1.2 10 mρπ−−×==××##4.#(b)(The#nuclear#density#of#an#alpha#particle#will#be#the#same#as#the#nuclear#density#of#thorium#because #the#nuclear#density#is#independent#of#the#mass#number.##5.((c)#Write#the#nuclear#density#of#an #alp h a#p artic le: #17 32.3 10 kg/mρ=×##Insight:#The#density#of#all#nuclei#is#approximately#2.3×1017#kg/m3,#regardl e ss #o f#t h e #identity#of#t h e #is o t o p e . #((13.##Picture(the(Problem:#A#nucleus#undergoes#α#decay.##Strategy:#The#radius#of#a#nucleus#depends#upon#the#total#number#of#nucleons#A#that#it#contains#(equation #32 H4).##Alpha#decay#involves#the#emission#of#four#nucleons#from#the#nucleus.##Use#these#facts#to#answer#the#conceptual#question.##Solution:#1.((a)#In#α#decay,#the#radius#of#the#daughter#nucleus#is# less#than#that#of#the #original#nucleus,#because#the#daughter#nucleus#contains#four#fewer#nucleons.###2.((b)#The#best#explanation#is#II. #When#the#nucleus#undergoes#decay#it#ejects#two#neutrons#and#two#protons.##This#decreases#the#number#of#nucleons#in#the#nucleus,#and#therefore#its#ra dius#will#decrease.##Statements#I#and#III#are#each#false.##Insight:#Statement#III#is#true#for#beta#decay#(β)#or#gamma#deca y#(γ),#but#not#alpha#decay.####14.##Picture(the(Problem:#A#nucleus#undergoes#β#decay.##Strategy:#The#radius#of#a#nucleus#depends#upon#the#total#number#of#nucleons#A#that#it#contains#(equation #32 H4).##Beta#decay#involves#the#emission#of#an#electron#from#the#nucleus,#but#the#number#of#nucleons#remains#unchanged.##Use#these#facts#to#answer#the#conceptual#question.##Solution:#1.((a)#In#β decay#the#radius#of#the#daughter#nucleus#is#the#sam e#as#that#of#the#original#nucleus,#because#the#daughter#nucleus#contains#the#same#number#of#nucleons.###2.((b)#The#best#explanation#is#III.#When#a#nucleus#emits#a# #particle#a#neutron#is#converted#to#a#proton,#but #the#number#of#nucleons#is#unchanged.##As#a#result,#the#rad ius# of#th e# daughter#nucleus#is#the #sam e#as#that#of#the#original#nucleus. ##Statements#I#and#II#are#each#false.##Insight:#In#refere nc e#to#state m en t#I,#it#is#possible#for#a#n uc leus #to#gain #an #electro n#in#a #proc ess#ca lled#elec tron #capture.##However,#even#in#that#process#the#number#of#nucleons#remains#the#same,#and#so#does#the#radius#of#the#nucleus.##(26.##Picture(the(Problem:#CarbonH14#undergoes#beta#decay#with#a#halfHlife#of#5730#y.##Strategy: Consider the random nature of radioactive decay when answering the conceptual questions. #Solution:(1.((a)(Yes,#it#is#possibl e#fo r#a #pa rt icu l ar #nu c le u s#in #a#s ample#of#carbo nH14#to#decay#after#only#1#s#has#passed#because#of#the#entirely#random#nature#of#radioactive#decay.##The#halfHlife#only#de sc ribe s#th e#a ve rag e#time#required#for#half#of#a#large#num b er#of#nuc lei#to#decay .##2.((b)(Yes,#it#is#possib l e#fo r#a #pa r ticu l ar #n uc le u s#in #a#s ample#of#carb onH14#to#decay#after#10,000#y#have#passed#because#of#the#entirely#random#nature#of#radioactive#decay.####Insight:#All#we#can#say#for#sure#is#that#half#of#the#initial#carbonH14#nuclei#(on#average)#will#have#decayed#after#5730#years#have#passed.##((31.##Picture(the(Problem:#The#number#of#radioactive#nuclei#in#a#sample#decays#to#a#sixteenth#of#its#initial#va lu e #within#a#specified#period#of#time.##Strategy: In each half-life the number of nuclei drops by a factor of two. Calculate how many half-lives are necessary to drop to a sixteenth of the initial number of nuclei. Divide the total time by the number of half-lives to calculate the half-life. #Solution:#1.(Calculate#the#number#of#halfHlives:(#11 4216nn⎛⎞=⇒=⎜⎟⎝⎠##2.#Divide#the#time#by#the#number#of#halfHlives:(#1218 d4.5 d4TTn== =##Insight:#Note#that#after#each#halfHlife#the#num ber#of#radioactive#nuclei#decreases#by#one#half.#Therefore,#after#4.5#days#1/2#are#left,#after#9.0#days#1/ 4#a re #left,#a fte r#13.5#da ys#1/ 8#are#left,#a n d
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