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  Nuclear Reactions, Physics tutorial

Introduction:

The main forces working inside a nucleus are because of the volume and surface term. This causes the nucleus to act like a liquid drop. Thus, if energy is functioned from the external sources, oscillation modes are excited in same manner as for a drop of liquid. In this reference, nucleuses can fission into two pieces that is the process considered as nuclear fission. Also, two or more light nuclei can blend together to make heavy nuclide.

Nuclear Fission:

When a nuclear fission occurs, two or more free neutrons are freed. Fission is similar to any other nuclear reactions in which total charges and total number of nucleus should be remaining constant. Example

235U92 + 1η0 → 236U92

→ A1 P1Z1 + A2 P2Z2  + 1 K0η      

Where K is number of neutrons emitted. Usually A1 ≠ A2.                  

741_Nuclear fission.jpg

From the diagram above A1  is about 95 and A2  is approx 140. From experiment result, average value of K from this reaction is 2.5.

Energy released from neutron yield

We can calculate the energy released as follows:

235U92 + 1n0 →(236U92) →98Zr40 →β- 98cf41→β-  98Mo40 + 21n0

→136Te52 → β-  136T53→ β-  136Xe54 + 21n0

235U92 + 1n0 →98Mo40 + 136Xe54 + 21n0 + 40e-1

Σmi = 236.133amu

Σmf = 235.905amu

Δm = Σmi - Σmf

= 0.228amu

= 0.228amu x 931.5MeV = 212.268MeV

212.268 x 1.6 x 10-13

3.3 x 10-11J

In addition to this energy, some energy has been carried away through released β- particles and rays. Fission processes happen sequentially taking out approximately 210MeV at each point of fission. So if fission is acceptable to continue, one can build a large amount of energy 3.36 x 10-11J in a single fission when this is multiplied by Avogadro's number, we get total energy free in 1 atomic mass that is in 235g of 235U        

= 3.36 x 10-11 x 6.02x 1023

= 2.02 x 1013J

If this is consumed as fuel in one month in a nuclear power reactor, the power output

 Pout  = 2.02 x 1013 J / 30 x 24 x 3600

= 7.8 x 106W

= 7.8 Me V

finally, this is more  than  energy desired by any University in  country.

Thermonuclear Reaction Or Nuclear Fusion

 Fusion is  synthesis of heavier nuclei from light ones and this can occur with  liberation of energy specially in cases where total mass of the product is less than  total mass of reactant  that is  implication Δm>0.

Σmf - Σmi = Δm>0

 This is always the case for which A1 + A2< 60 example:

1H1 + 1H1 → 2H1 + β+ + 0.4MeV

The reactants have to known high kinetic energy to overcome coulomb's repulsive form among them, but nuclear force of attraction takes over to fuse them together. In other to produce this kinetic energy, temperature of particles is increased and energy is produced through thermal agitation example. To generate 480KeV thermally, the temperature must be approximately 3.7x109K. This is why reaction is considered as thermonuclear reaction. Because of this high temperature requirement, no nuclear power plant is depending on nuclear fusion. Thermonuclear reaction can be acheived by fusing proton and deuteron together

(1H1  & 2 H1),( 1H1  & 1H1),( 2 H1  & 2 H1) etc

The reaction involving the fusion of deuteron and tritium

 2 H1  +  3 H1  →4 He2 + 1n0 + 17.6MeV              

This reaction yields a huge amount of energy and dissipates heat in inconceivable short time of 1.2x10-6s, hence releasing extremely high level of power. This is base of the "modern hydrogen bomb". Fusion of light elements is recognized as source of "stellar energy". In interior of sun, temperature is high sufficient for nuclear fusion to take place.

Criticality of a Reactor:

From diffusion equation

D∇2Φ - ΣaΦ + S = 0

Σa = Σf + Σac

Number of neutrons which will be absorbed by fuel would be provided as

Σa = Σf + Σac and not all neutrons absorbed by fuel lead to fission. If η is probability of producing more neutron or is average no of neutron emitted per neutron absorbed. Thus total no of neutron emitted by fuel is given as = ηΣafΦ

Equation can further be written as:

ηΣa(Σaf/Σa)Φ

ηfΣaΦ

This signifies that

f = Σaf/Σa = Utilization factor

Σaf is absorption cross section by the fuel.

Σa is total absorption cross section.

Ratio between numbers of neutrons emitted to number of neutrons absorbed is provided as:

ηfΣaΦ/ΣaΦ = K∞

As number of neutrons absorbed = ηΣaΦ: Thus total number of neutrons emitted and absorbed by fuel in second medium is given as:

ηfΣaΦ x Σaf = ηfΣafΣaΦ

For total of number of neutrons emitted by fuel

ηx(ηfΣafΣaΦ) = η2fΣafΣaΦ

From equations we can write:

ηfΣaΦ = K∞ΣaΦ

Thus, diffusion equation can be written as:

D∇2Φ - ΣaΦ + K∞ΣaΦ = 0

Dividing through by D

∇2Φ - 1/L2Φ + K∞Φ/L2 = 0

∇2Φ + (K∞-1/L2)Φ = 0

Put (K∞-1/L2) as B2

∇2Φ + B2Φ = 0

B is known as buckling factor. In system where K∞>1, system is said to be Super critical

K∞ = 1 system is said to be Critical

K∞ < 1 the system is said to be sub critical

Value of K∞ can be controlled in the reactor e.g. to increase power in reactor, K∞ will be increased. Value of K∞ can't be controlled under explosive and nuclear reaction. Control rods like carbon rods are utilized to control K in system or reactor.

If K∞ is greater than 1, it signifies number of fission increases from generation. In this situation, energy released by chain reaction increases with time hence chain reaction. System is said to be Super critical: if K∞< 1, number of fission decreases with time and chain reaction said to be sub critical. If K∞ = 1, chain reaction proceed at the constant rate, energy is released at the steady level, system is termed critical. To increase power being generated by reactor, operator increases K to value greater than unity so that reactor becomes super critical. When desired power level has been achieved, it returns reactor to the critical by altering valued of K to be unity and reactor then maintains specified power level.

To reduce power or shut reactor down, the operator just reduces K1 making reactor sub-critical and consequently, output power at the system decreases.          

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