202 CHAPTER12. THEHYDROGENATOM
becausethevolumeofasoliddependsmainlyonthenumberofatoms,andthesize
ofeachatom. Theenergyreleasedinachemicalreactionalsodependsonthenumber
ofmoleculesparticipatinginthereaction,andtheenergyreleasedbyeachmolecule.
ThenumberofmoleculesinamoleisknownasAvogadro’snumber,whichis
NA = 6. 02 × 1023≈1 gram
mprotoningrams(12.55)
The protonmassgives slightlythe wrong value, due to the proton-neutron mass
difference. Anyway,NAtellsusapproximatelyhowmanyprotonsorneutronsthere
areinagram,orhowmanymoleculesinamole(=(no. ofprotons+neutronsper
molecule)×onegram).Quantummechanicsthenpredicts,inprinciple,thevolume
ofeachmolecule,andtheenergyreleasedinanygivenchemicalreaction.
Ofcourse, usingquantum mechanics to find the precise volume of any given
moleculeisaformidabletask. Butwecanuse thesolutionof thehydrogenatom
tomakeorder-of-magnitudeestimates.Thevolumeofanygivenmoleculeiscertainly
largerthanthatofthehydrogenatom,butusuallynotbymuchmorethanoneor
twopowersof10.Likewise,theenergyreleasedinachemicalreactionistypicallyless
thanthehydrogenatombindingenergy,butthisisstillaroughorder-of-magnitude
estimate. Ifwearesatisfiedwithaccuracywithinacoupleofordersofmagnitude,
thenwecananswerthetwoquestionsposedabove:
- Whatisthevolumeofamoleof(solidorliquid)anything?
- Howmuchenergyisreleasedbyburningamoleofanything?
Startwiththefirstquestion. Wewillimaginethatallmoleculesfitinsideacube
oflengthequaltothediameteroftheHydrogenatominitsgroundstate.Thisgives
usavolume
v∼(2a 0 )^3 (12.56)
wherea 0 istheBohrradius
a 0 =h ̄^2
me^2= 5 × 10 −^9 cm (12.57)sothat
v∼ 10 −^24 cm^3 (12.58)
Thenthevolumeofamoleis
Vmole=NAv∼ 0. 6 cm^3 (12.59)whichisatleasttherightorderofmagnitude. Thevolumeofamoleofanythingis
afewcubiccentimeters,notcubickilometersorcubicparsecs.