qsp_chapter7-boltzmanasd

Chapter 7

Applications of theBoltzmannDistribution inClassical Physics

Topics

Examples of the use of the Boltzmann distribution in classical physics. Sedimenta-tion. Chemical reactions. Phase equilibrium and vapour pressure. The Origin ofhelium in the Big Bang.

In this chapter, we apply the Boltzmann distribu-tion to a variety of different problems in classicalphysics.

7.1 Sedimentation

A pleasant and important application of the Boltz-mann distribution is to the phenomenon of the sed-imentation of small particles in a fluid. We demon-strated the random motions of microspheres dif-fusing in distilled water as an example of the phe-nomenon of Brownian motion. The path was as-sociated with the vast numbers of random encoun-ters of water molecules with the microspheres. Aswe showed, the displacement of the particles wasexactly as predicted by Einsteins theory of theprocess. The same process of molecular collisions

1

The Classical Boltzmann Distribution 2

means that the tiny particles do not just settle atthe bottom of a test-tube, but form an atmospherein the test-tube, just like that of molecules in theEarths atmosphere.

As in the case of the Earths atmosphere, we canwork out how the potential energies of the micro-scopic particles vary with height within the test-tube. We suppose that the particles have densityp and the fluid density f . The net force act-ing downwards is the difference between the down-ward force of gravity mg and the upthrust u due tothe displaced fluid, that is, Archimedes principle.

Archimedes PrincipleThe upthrust is equal to the weight of dis-placed fluid.Therefore, the net downward force is

f = mgu = mgf mpg = mg

(1 f

p

), (7.1)

where m/p is the volume of the particle. There-fore, the potential energy at height h from the bot-tom of the test-tube is

(h) = h0mg

(1 f

p

)dh = mgh

(1 f

p

).

(7.2)Therefore, just like the Earths atmosphere, the dis-tribution of particles with height is expected to be

n(h) = n(0) exp[(h)

kT

],

= n(0) exp[mgh

kT

(1 f

p

)], (7.3)

= n(0) exp( hh0

), (7.4)

where h0 = (kT/mg)[1 (f/p)]1. This was his-Figure 7.1. Perrins results for the height

distribution of small particles in a test-tube.The vertical axis is plotted on a logarithmic

scale.

torically an important calculation because Perrinvalidated this analysis by making observations ofthe height distribution of tiny particles and foundthe distribution shown in Figure 7.1. At a partic-ularly fraught period in the history of physics, thiswas important evidence in favour of the molecularpicture of the internal structure of matter. It canbe seen from (7.3) that, if the sizes and masses ofthe particles are known, Boltzmanns constant kcan be estimated.

The process of sedimentation can be used to sep-arate out large molecules in what is known as an


ultracentrifuge. The lowest mass particles have thelargest atmospheric scale heights. With accelera-tions of 106g, macromolecules such as proteins canbe separated out.

7.2 Chemical Reactions

The Boltzmann distribution finds wide applicationin chemistry, particularly in understanding the rateat which chemical reactions take place. Typically,chemists mix two chemicals together so that theproducts of the reaction are in lower energy statesthan those of the reactants. In order for a reactionsuch as

A + B Cto take place, two things must happen:

The molecules A and B must collide; In order to react, they must overcome repul-sive forces which lead to a potential energyor activation energy barrier, as illustrated inFigure 7.2.

Very often, it is a good idea to heat up the solutionto speed up the reaction. Why does this help?

Figure 7.2. The potential energy associatedwith the atoms involved in a

chemical reaction.

|f|

|b|

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PotentialEnergy

Product

Reactants

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The diagram shows the potential barrier f whichhas to be overcome before the reactants A and Bcan combine to reach a lower energy state in form-ing the product C. Equally, the reverse dissociationreaction

C A+ Bhas a greater activation energy b.

In chapter 5, we showed that the number of colli-sions, N , which a particle makes in travelling adistance l is

N = nl,

where n is the number density of molecules and is the collision cross-section. If the particles havespeed v, l vt and therefore the rate of colli-sions is

dNdt

= nv. (7.5)


The origin of the activation energy is the kineticenergy of the colliding molecules. To make a sim-ple estimate of what happens, let us assume thatmolecule A collides with a stationary molecule B.In order for the reaction to take place, the kineticenergy of molecule A must be at least 12mv

2A = f

and so we need to estimate the fraction of atomswhich have kinetic energies greater than this value.

Reminder - The Three-dimensionalMaxwell Distribution

f(v) dv =( m2pikT

)3/24piv2 emv

2/2kTdvAs shown in (7.5), the probability of a collision isproportional to the speed v of the particles and sothe fraction of atoms colliding with speed greaterthan vf =

2f/m is

=

2f/m

v v2 exp(mv

2

2kT

)dv

0v v2 exp

(mv

2

2kT

)dv

(7.6)

= exp(f/kT )( fkT

+ 1), (7.7)

exp(f/kT ), if f kT.Optional practice in integrationShow that the ratio of the integrals in (7.6)gives the result (7.7).

This is a rather crude estimate and can be improvedin many ways, but the calculations are lengthy. Thefinal result is just the one we have derived

= exp(f/kT ), (7.8)which is again the simple form of the Boltzmannfactor.

We need also to include the fact that, for each atomA, the collision rate is proportional to the numberdensity of atoms B, nB and that the number densityof A atoms is nA. Therefore the overall rate of thereaction must also be proportional to nAnB and so

dnCdt

nAnB exp( fkT

)(7.9)

Those of you doing chemistry will know all aboutthis. The importance of heating up the reactants isthat T is raised and so the probability of the react-ing particles having enough energy to pass over theactivation barrier increases exponentially, as f/kTdecreases.


Notice the other interesting point that the dissocia-tion reaction is very much less likely to occur sinceb/kT f/kT and these terms appear in the ex-ponential at the same temperature T . Notice alsothat the energy liberated in the reaction is b f .This formula can be tested by plotting the log-arithm of the reaction rate against 1/T , what isknown as an Arrhenius plot. An example of such aplot for the decomposition of acetaldehyde is shownin Fig. 7.3.

Figure 7.3. Arrhenius plot for the reactionrate of the decomposition of acetaldehyde.

This diagram illustrates the problems facing deter-gent manufacturers. One of the big challenges is todiscover detergents which work at lower tempera-tures so that less energy is used in heating up thewater and there is less chance of being scalded. Thebig problem they have is to beat the effects of theexponential in the Boltzmann distribution.

7.3 Phase Equilibrium and Vapour Pres-sure

7.3.1 The Structure of Solids and Liquids

Let us make a brief excursion into the structuresof solids and liquids in order to understand thephysics of melting and boiling. We represent theparticles by hard elastic spheres. In the case ofgases, we have assumed that interatomic or inter-molecular forces can be neglected, but that cannotbe the case for liquids and solids, which are heldtogether by the attractive forces between particles.For simplicity, we assume that the force betweenparticles is represented by the potential function(r/a0) shown in Figure 7.4, in which r is the sep-aration of the particles and a0 is their equilibriumseparation.

Figure 7.4. The attractive potentialbetween two atoms or molecules.

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.........

.........

.........

.......

...........0

1

0.5

0.5

1

1 2 3r/a0

(r

a0

)

a0 ||0||

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When the atoms are far apart, the negative poten-tial means that they are attracted to each other,whilst at short distances they are repelled by quan-tum mechanical forces. The equilibrium separationoccurs at some separation a0, at which the attrac-tive potential has a minimum of depth 0. 0 isthe amount of energy which would have to be sup-


plied to dissociate the molecule, that is, separatethe atoms to infinity.

If the particles were spheres, the closest way ofpacking them together in a solid is as a close packedarray. This is the type of arrangement you find ifyou build a pyramid of identical oranges or billiardballs. A close-packed plane is shown in Figure 7.5.The spheres are packed as close together as they

Figure 7.5. A Close-packed Plane

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can be in a plane. Their centres all form equilat-eral triangles so that, in the plane, any atom issurrounded by six nearest neighbours. The inter-atomic forces keep all the atoms at the same dis-tance apart.

Now, we have to place layers of close-packed planesone top of one another. The closest way of packingthem in three-dimensions is to place a sphere ontop of each triangle formed by a triplet of atoms,as shown in figure 7.6 for a pair of rows. Thus,

Figure 7.6. A Close-packed Planewith the Next Row on Top

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in a close-packed array, the atoms form a continu-ous sequence of tetrahedra. When we place a sec-ond plane on top of the first as shown, and then athird on top of that, we can count up the numberof atoms which are at the same distance from anyatom. In the plane, there are 6 atoms at the samedistance. In the upper plane, there are three atthe same distance as those in the central plane andthe same goes for the atoms in the lower plane.Therefore, each atom is surrounded by 12 atoms,all at the same distance. This is known as a close-packed array and we say that the coordination num-ber N = 12. The array forms a crystal lattice withlong-range crystalline order.

The above picture represents a close-packed solidat T = 0 K. As the temperature is raised aboveT = 0, if we consider only the translational mo-tions of the atoms of the array, they will acquirea kinetic energy of 32kT per atom and so they willrandomly bang around within a small volume, orcage, defined by the 12 nearest neighbours. If thetemperature is not too high, the crystal structureis not destroyed.

As the temperature is raised, however, the randomthermal motions make the lattice expand, and, ifthis is not prevented by the application of an in-


creased external pressure, the molecules can movesufficiently far apart for occasional rearrangementsof the crystal lattice to occur. If sufficient rear-rangements take place, the long-range crystallineorder is lost and the material become a liquid. Apartfrom the fact that these rearrangements can takeplace, the situation is not so different from that ofa solid. Each molecule still bangs around in a cageformed by its nearest neighbours, but the structureof the cage is no longer a regular crystalline lattice.

Figure 7.7. Random Close-packingin a Liquid

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For a liquid in the low temperature limit, this ar-rangement corresponds to random close packing, asillustrated in Figure 7.7. This structure is obtainedif we stick the spheres together at the hard sphereseparation, but at random orientations. The near-est neighbour distance remains the same as in thesolid. There is some short-range order, but thereis no longer any large-scale crystalline order. Forrandom close-packing, the coordination number isN 10, rather than 12 for the close-packed array.We would therefore expect liquids to be typicallyabout 20% less dense than solids.

7.3.2 The Binding Energies of Solids and Liq-uids

By binding energy, we mean the energy which hasto be supplied in order to break up the solid intoits component atoms or molecules at T = 0. Fromthe shape of the inter-particle potential, it can beseen that the magnitude of the interaction poten-tial energy decreases rapidly with increasing dis-tance from a0 and so we need only take nearestneighbour interactions into account. We can there-fore estimate the binding energy by noting that,according to the close-packed model, most of thebinding energy of a molecule is with its 12 nearestneighbours. Now, the potential refers to the inter-action energy of pairs of molecules and thereforewe share half the potential energy between each ofthem. Therefore, the total energy needed to dis-perse the molecules to infinity is 6NA0 per mole,that is,

Binding energy of solid = 6NA0. (7.10)


We can apply the same line of reasoning to thebinding energies of liquids. The only difference isthat there are only 10 nearest neighbours, each sep-arated by a0. Again pairs of particles share thebinding energy pair-wise and so the total bindingenergy per mole necessary to disperse the fluid toinfinity is

Binding energy of liquid = 5NA0. (7.11)

7.3.3 The Latent Heats of Solids and Liq-uids

Substances can exist as solids, liquids or gases (orvapours) and these are known as their differentphases. The processes by which substances changephase are:

Melting solid liquid Solidification liquid solid.Boiling liquid gas Condensation gas liquid.Sublimation solid gas Condensation gas solid.At a given pressure, these phase transitions takeplace at a particular temperature. The processeson the left of the list involve the absorption of heat;those on the right involve the release of energyas the substance reaches a state of lower internalenergy. The energies associated with these phasetransitions are known as their latent heats.

We can estimate the latent heats associated withthese processes by arguments based upon the prin-ciple of the conservation of energy. We assume thatthe transitions take place at a particular tempera-ture and then estimate the total internal energiesin the initial and final states of the phase transi-tion. To do this, we need to estimate the internalenergies U of the substance in both phases.

Let us give an elementary treatment in terms ofthe internal binding energies of the substance andignore the energies associated with the thermal mo-tions of the particles. To a good approximation, theinternal energy in the solid phase is 6NA0 and5NA0 in the liquid phase. In the gaseous phase,since we are ignoring thermal motions, we set theinternal energy equal to zero.


Thus, the molar latent heat for the transition fromthe solid to the gaseous phase, a sublimation phasetransition, is

Lsublimation 6NA0. (7.12)

A similar line of reasoning leads to an estimate ofthe latent heat associated with the liquid to gastransition. Hence the latent heat of boiling is

Lboiling 5NA0. (7.13)

By subtraction, we can find the latent heat of melt-ing.

Lmelting NA0. (7.14)Thus, melting is a much less energetic process thanboiling this explains why scalding is so painful.

You should be aware that these are simple esti-mates, but they work well, provided they are notpushed too far.

7.3.4 Boiling water

In equilibrium, molecules leave the surface of a liq-uid if they can overcome the potential barrier asso-ciated with their attraction for the bulk of the fluid.At the same time, molecules in the vapour phasecan be absorbed into the liquid. Let us consider theequilibrium state within a closed vessel containinga liquid (Figure 7.8).

Figure 7.8. A Liquid in Equilibriumwith its Vapour

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The simplest model is to consider an atom or moleculelocated at the surface. If it were within the body ofthe fluid, it would have on average 10 bonds linkingit to the bulk of the fluid. At the surface it has onlyhalf the number of bonds, each with energy 0.Therefore, we would expect that the atom wouldhave to acquire an energy = 50 = L/NA fromrandom thermal motions within the fluid in orderto escape from the surface. We recognise that 50is just the latent heat per molecule. This is whatis known as a thermally activated process.

The distribution of particle speeds perpendicularto the surface of the liquid is given by the one-


dimensional Maxwell distribution

dn(vz) = nLf1(vz) dvz = nL( m2pikT

)1/2exp

(mv

2z

2kT

)dvz.

(7.15)The rate of arrival of particles with speeds vz tovz + dvz at the surface is vz dn (vz) particles m2

and so the rate of escape of molecules from thesurface with energies greater that is

n( ) = 2/m

vz dn (vz)

= nL( m2pikT

)1/2 2/m

vz exp(mv

2z

2kT

)dvz,

= nL

(kT

2pim

)1/2exp

( kT

)(7.16)

In equilibrium, the loss of particles from the surfaceof the liquid is balanced by the return of particlesfrom the vapour to the liquid state. The rate ofarrival of particles from the vapour phase to thesurface is given by the usual formula 14nVv, wherev =

8kT/pim and nV is the number density of

particles in the vapour phase. Therefore, when theprocesses of escape and return are in balance, weexpect

nL

(kT

2pim

)1/2exp

( kT

)= 14nV

(8kTpim

)1/2,

(7.17)that is,

nVnL

= exp( kT

)= exp

( LRT

). (7.18)

We see that we have recovered another expressioninvolving the Boltzmann factor. What this expres-sion means is that the atoms or molecules in thevapour phase have average energy greater thanthose in the liquid phase and this makes sense sincethey must have at least energy to escape from thesurface of the liquid.

Now, as we change the temperature, the numberdensity of atoms in the liquid phase scarcely changes


at all and so the pressure of the vapour in equilib-rium with the liquid is

p = nVkT T exp( LRT

). (7.19)

The variation of the vapour pressure with tempera-ture is dominated by the exponential term. We see Exercise Check this calculation for your-

self, given that the latent heat of vapori-sation of water is L = 4.06 104 J mol1.

that, as the temperature changes from 0 to 30C(273 to 303 K), the pre-exponential factor changesby a factor of 1.1, while the exponential changes bya factor of about 6.

Figure 7.9 shows how the vapour pressure changeswith temperature for water. Often, the term satu-rated vapour pressure is used for the vapour pres-sure since the pressure is as great as it can be atthat temperature. When the pressure of the vapourbecomes equal to local atmospheric pressure, theliquid boils. It can therefore be understood fromFigure 7.9 why the boiling point of water decreasesat high altitude. The saturated vapour pressure ofthe water vapour reaches local atmospheric pres-sure at a lower temperature.

7.4 The Origin of Helium - Non-examinable

Let us complete this study by working out the basicphysics of the Big Bang and, in particular, the ori-gin of helium. Everywhere we look in the Universe,the abundance of helium, 4He, is at least 23% bymass. This is too large a percentage to have beenproduced by nucleosynthesis in stars, but it comesnaturally out of the early phases of the Big Bang.This important result depends only upon funda-mental physics and the Boltzmann distribution.

Figure 7.9. The variation of vapour pressureof water with temperature. Notice that the

horizontal axis shows 1000T1.

First, we need the thermal history of the Universe.This turns out to be remarkably simple. The cooledremnant of the hot early phases of the Big Bangare all around us as microwave background radia-tion. This radiation is incredibly uniform over thesky and has a perfect black body spectrum at atemperature of 2.726 K (Figure 7.10). Now, theUniverse is expanding uniformly on the very largescale and so we can work out how the tempera-ture and energy density of the radiation change as


the Universe expands. In fact, the radiation simplycools adiabatically as the Universe expands. Let uswork out the consequences of this expansion for thetemperature history of the Universe.

Figure 7.10. The spectrum of the CosmicMicrowave Background Radiation asmeasured by the COBE satellite. The

spectrum is that of a perfect black-body at atemperature of 2.726 K.

7.4.1 The Equation of State for a Photon Gas

Just as we can write p = nkT for a perfect gas, sowe can find a relation between pressure and temper-ature for a gas of electromagnetic waves or photons.Let us do this as an example.

Example: Show that the equation of state of aphoton gas is p = 13u =

13aT

4, where u is the en-ergy denisty of the radiation.

First, we ask,

What is the probability of particles havingvelocity vectors pointing at an angle withrespect to a chosen direction?

We soon realise that there will be none coming in atprecisely . Rather, we should have asked,

How many particles are there with velocityvectors pointing between the angles and+d with respect to the chosen direction?

Remember that we can make d as tiny as we like. Theway of working out the probability is shown in Figure7.11.

Figure 7.11. Probability Distributionson a Sphere

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r

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.

0

r sin

d

dl

Because of the effect of collisions, the velocity vectorsof the particles are randomised and so there is an equalprobability of the velocity vectors pointing in any direc-tion. Therefore, we can consider the probability of theparticles having velocity vectors lying within some solidangle d in terms of the element of area dA on a spherein that direction (see Section 9 of the Maths Handbook).Thus, the probability is the ratio of the area of dA tothe total surface area of the sphere

p(dA) =dA4pir2

(7.20)

This result is quite general for an isotropic distributionof directions. In our case, we want to know the probabil-ity of the particles having velocity vectors within angles


to + d of our chosen direction. It can be seen fromthe diagram that the tips of the vectors could lie any-where within the circular annulus shown on the sphereand they would satisfy that criterion. Thus, we need towork out the surface area of that annulus and insert itinto (7.20). Since the annulus is very thin, we simplymultiply the circumference of the annulus by its thick-ness dl, as shown on the diagram. The radius of theannulus is r sin and so its circumference is 2pir sin .The thickness of the annulus is dl = r d. Therefore,its surface area is dA = 2pir sin r d = 2pir2 sin d.The probability of the velocity vectors lying within theangles to + d with respect to the chosen directionis therefore

p(dA) =dA4pir2

= p() d =2pir2 sin d

4pir2= 12 sin d

(7.21)It immediately follows that, if we have N particles, thenumber with velocity vectors between angles to +dis

N() d = Np() d = 12N sin d (7.22)

Next, we need the flux of particles arriving from a par-ticular direction, that is, the number of particles movingwith velocity v through an elementary of surface area dAper second. We recall how we define the vector area dA(see Section 6 of the Maths Handbook). The numberdensity of the particles is n.

Figure 7.12. The Flux of Particles though anelementary area dA

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v

dA

v

dA

S

The component of velocity parallel to the plane of theelementary area is |v| sin , but this results in no flowthrough the area dA. The component of velocity paral-lel to the vector dA, that is, perpendicular to the sur-face, is |v| cos and so the number of particles passingthrough dA per second is n|v||dA| cos = nv dA (Fig-ure 7.12).

We now combine these results to relate the pressure tothe energy density of a photon gas. Suppose the numberdensity of photons is n. Consider those photons propa-gating towards the wall of the vessel at angle of incidence. Then, the change in momentum when the photon isreflected from the wall is 2h cos /c, using p = h/c.The flux of such photons is nv dA = nc cos dA. Thenumber of photons arriving within angle to + d is12N sin d and so the rate of change of momentum per


unit area, or the pressure of the gas of photons, is

p = pi/20

2h cos c

nc cos 12 sin d,

= nh pi/20

cos2 d(cos ),

= 13nh =13u. (7.23)

Hence, for isotropic radiation in general and black-body radiation in particular, we find p = 13u.

The equation of state of a photon gas

p = 13u.Now, let us use this result to work out the adiabaticexpansion of a spherical volume V with radius R fora gas of photons. The system is thermally isolatedand so

dU = pdV. (7.24)For the photon gas, we need an expression for theenergy density of radiation at a given temperaturein thermal equilibrium. This is given by the Stefan-Boltzmann law, u = aT 4, where a is known as theStefan-Boltzmann constant we will study this lawin detail later in the course. Therefore, the totalenergy density is U = V aT 4, and p = 13aT

4. Sub-

The Stefan-Boltzmann LawThe energy density of radiation in thermalequilibrium at temperature T is

u = aT 4,

where a = 7.566 1016 J m3 K4 isknown as the Stefan-Boltzmann constant.This law will be discussed in detail in thequantum physics half of the course. It canbe found by integrating the energy underthe black-body curve, which was quotedin Figure 1.11.

stituting into (7.24), we find

dU = p dV,d(V aT 4) = 13aT 4 dV,

4V T 3 dT + T 4 dV =T 4

3dV,

4V T 3 dT = 43T 4 dV,

dTT

= 13dVV

,

lnT = 13lnV,

T V 1/3 R1. (7.25)

We have implicitly worked out the ratio of specificheats for a photon gas. You will recall that weshowed that the expression (6.20) for the adiabaticexpansion of any gas was

TV 1 = constant.

It follows immediately that = 4/3 for a photongas.

Ratio of specific heats for a photongas

=43.


Now, we can use (7.25) to work out the temperatureof the radiation when it was squashed much closertogether than it is today. When we work out thedynamics of the expanding Universe, it turns outthat the thermal history of the early Universe isgiven by

T 1010 t1/2 K,so that, when the Universe was only 1 second old,the temperature was 1010 K. Since T R1, theUniverse was squashed by a factor of 1010/2.726 3.5 109 relative to its present size. This is sucha high temperature that all nuclei were broken upinto their constituent protons and neutrons. Atearlier times, the Universe was a sea of elemen-tary particles which were in thermal equilibrium ata very high temperature, which continued to in-crease, roughly as (7.25) into the very early Uni-verse. This is why the particle physicists regardthe very early Universe as a laboratory for particlephysics.

Let us now run the clocks forward from an equilib-rium state when the temperature was about 1011

K. At that stage, the electrons e, positrons e+,neutrons n, protons p, electron neutrinos e andantineutrinos e were all present in their equilib-rium abundances. The reactions which maintainthe protons and neutrons in thermal equilibriumare the following weak interactions

e+ + n p + e e + n p + e (7.26)

Now, the equilibrium abundances of neutrons andprotons is given by the standard Boltzmann rela-tion

N exp(E

kT

)(7.27)

where E = mc2 is the mass of the neutron or theproton. Since the mass of the neutron is slightlygreater than the mass of the proton,mn = 1.6749291027 kg and mp = 1.672623 1027 kg, the ratioof abundances of neutrons to protons decreases as


the temperature drops

NnNp

= exp(mnc

2

kT

)exp

(mpc

2

kT

)(7.28)

= exp(mc

2

kT

)(7.29)

where mc2 = (mn mp)c2 = 1.28 MeV is themass difference between the neutron and the pro-ton.

The relative abundances of neutrons to protons isdescribed by this formula until the time when thetime-scale for the weak interactions described by(7.26) becomes greater than the age of the Uni-verse at these early times. Detailed calculationsshow that this occurred when the Universe wasabout 1 second old, when the temperature of theUniverse was T = 1010 K or kT = 1 MeV. There-fore, inserting this value into (7.29), the neutron toproton ratio was frozen at a value of roughly 0.21.In the subsequent reactions, the neutrons combine

Figure 7.13. The predicted primordialabundances of the light elements created

during the early phases of the Big Bang as afunction of the present baryon density of the

Universe.

with protons to form helium nuclei by a sequenceof reactions in which first deuterium is formed bythe combination of a neutron and a proton. Thedeuterons then combine with protons to form 3Hewhich then combine with other 3He nuclei to form4He. A more detailed calculation shows that about24% by mass of 4He is formed by this process Figure7.13). Remarkably, this is precisely the minimumvalue of the helium abundance we find wherever wecan detect helium in the Universe. Thus, the abun-dance of helium acts as a thermometer to tell us thetemperature at which the neutrons and protons de-coupled in the very early Universe. In addition, theabundances of the other light elements such as deu-terium (D), helium-3 (3He) and lithium-7 (7Li) canbe accounted for by the same process of primordialnucleosynthesis.

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