ALL ABOUT NUCLEAR CROSSSECTION

The probability of a Nuclear Reaction can be defined in terms of number of particles emitted or number of nuclei undergoing transmutation for a specified number of incident particles.

It is usually expressed in terms of an effective area presented by a Nucleus towards the beam of bombarding particles, such that the number of incident particles that would strike such an area, calculated upon a purely geometrical basis, is the number observed to lead to Nuclear Reaction given in question.

This effective area is called crosssection for that reaction.

Thus the probability of occurrence of a particular Nuclear Reaction is described by effective crosssection for that process.

The crosssection may also be defined as 

1) The probability that an event may occur when a single nucleus is exposed to a beam of  particles of total flux one particle per unit area.

2) The probability that an event may occur when a single particle is shot perpendicularly at a target consisting of one particle per unit area.

The idea of crosssection gives imaginary area associated with each nucleus, the area is so chosen that if bombarded  particle passes through it the reaction takes place, otherwise it is not.

The total nuclear crosssection is effective area possessed by a nucleus for removing incident particles from a collimated beam by all possible process.

This can be written as sum of several partial crosssections which represent contributions to various distinct, independent processes which can remove particles  from incident beam.

Thus,

𝛔t  = 𝛔s  + 𝛔r                                                    ---------------(1)

𝛔t  is "Total crosssection"

𝛔s is "Scattering crosssection"

𝛔r is "Reaction Crossection"

Scattering Crossection

Scattering crosssection can be classified as 

i) Inelastic scattering

ii) Elastic Scattering

Thus, we get

𝛔s  = 𝛔el  + 𝛔inel                                                          --------(2)

These partial crossections can still be subdivided.

In case of elastic scattering separate partial crosssections cannot be written because of possibility of interference between them.

On other hand all inelastic scattering processses are incoherent and their crosssections are additive.  

𝛔inel  = 𝛔1  + 𝛔2 + 𝛔3 + .........                                      ------(3)

Differential crosssection

The distribution in angle of emitted particles in a nuclear reaction can be described in terms of a crosssection which is a function of angular coordinates in problem.

The crosssection which defines a distribution of emitted particles with respect to solid angle is called differential crosssection. It is defined by  d𝛔/dΩ.

Partial crosssection for a given process is

𝛔  = ∫(d𝛔/dΩ)*dΩ                                                                ------(4)

Expression of crosssection for a Nuclear Reaction

Consider a mono energetic beam of particles incident on a target shown in Fig.

Let the beam be uniform and contain ‘n’ particles per unit volume moving with a velocity ‘V’ with respective to stationary target.

Clearly the product ‘nV’ gives number of particles crossing a unit area perpendicular to beam per unit time. It defines flux ‘F’ of particles in incident beam.

                                            F = nV -------------------------------(1)

It is customary to normalize number of particles to one particle per volume ‘V’.

                                            n = 1/V ------------------------------(2)

The detector detects all particles scattered through an angle ‘𝛳' into solid angle dΩ.

The number of particles dN detected per unit time depends on following factors:

i)                    Flux of incident beam, F

ii)                   The solid angle, dΩ

iii)                 Number of independent scattering centers in target that are intercepted by the beam. Let these be N.

                                   dN = 𝛔(𝛳)*F*dΩ*N ---------------------(3)

𝛔(𝛳) is constant of proportionality defines differential scattering crosssection.

We can put

                                   𝛔(𝛳)*dΩ =d𝛔(𝛳) 

                       𝛔(𝛳) =d𝛔(𝛳) / dΩ -----------------------(4)

  The total number of particles scattered per unit time is obtained by integrating over  entire solid angle.

                                       N = F*N*𝛔total  --------------------------(5)

where,

    Total Crosssection  𝛔total  = ∫𝛔(𝛳) dΩ ---------------------(6)

 𝛔total has dimensions of area.

Unit used to express crosssections is barn.

1 barn = 10⁻²⁸ cm².

The area 'a' intercepted by beam contains 'N' scattering centers. Total number of incident particles per unit time is given by

Nincident = F*a, where 'a' is area intercepted by beam; 'F' is incident flux.

Total number of scattered particles per unit time is

Nscattered = F*N*𝛔total

(Nscattered/Nincident) = (N*𝛔total)/a   ------------------------(7)

𝛔total is equal to area effective in scattering for one scattering center.

What is Penetration depth in Super Conductors

 In 1935, F. London and H. London described Meissner effect and zero resistivity by adding two conditions E=0(absence of Resistivity) and B=0(Meissner effect) to Maxwells Electromagnetic equations.

According to them, the applied field does not suddenly drop to zero at the surface of super conductor, but decays exponentially according to equation

B(x) = Bₐexp(-x/ƛ)

B(x) - Magnetic field at depth 'x' of material

 Bₐ  - Applied field; ƛ - penetration depth

Penetration depth is the length or depth from surface of metal at which the magentic field falls to 1/e of its original value.

Generally the magnetic field is likely to penetrate a superconductor to a depth of 10 - 100 nm.

Penetration depth doesn't have a fixed value but varies with temperature.

ƛ = ƛ / [1-(T/Tc)⁴]

ALPHA PARTICLE SPECTRUM? - DETAILED EXPLANATION

We have discussed that every alpha(𝛼) emitter has only one associated 𝛼 energy. This is experimentally true for many 𝛼 emitter where one finds that velocity spectrum of alpha particle from these isotopes is always a shar line spectrum. This is to be expected that since the emission of an alpha particle is a result of energy transition between two different nuclear states.

in 1930, S. Rosenblum, in France proved by means of spectograph that the 𝛼 particles from Thc(Bi²¹²), all of which had been thought to have same energy, actually were consisted of number of groups of particles with slightly different energies. The 𝛼 particles form a given radioactive substance were collimated with slits and after deflection thru 180⁰ with strong magnetic field, formed lines on photographic plate.

For example, ₈₄Po²¹⁴ decays to ₈₂Pb²¹⁰ byemitting 4 groups of  𝛼 particles having ranges in air 6.91cm, 7.79 cm, 9.04cm and 11.51cm. These ranges correspond to energies 7.68MeV, 8.28MeV, 9.07MeV & 10.51 MeV respectively.

Another example is decay of ₉₀Th²²⁸. 

₉₀Th²²⁸     →    ₈₈Ra²²⁴ + 𝛼

It comprises of five groups of 𝛼-particles with different energies.

𝛼-particle which is emitted by transformation of excited state of parent nucleus to ground state of daughter nucleus will have maximum energy.

In the above example, out of 5, four groups of  𝛼-particles leave the daughter nuclei in excited state. The fifth group of 5.42MeV 𝛼-rays take one to ground state of daughter nuclei. We can note from figure that excited states of nuclei reach ground state by emitting 𝜸-rays shown by vertical wavy lines.

Thus the fine structure of 𝛼-spectrum tells us about energy levels in daughter nuclei. We emphasize that existence of these different 𝛼-energy groups and 𝜸-rays proves the existence of nuclear energy levels.

Long Range 𝛼-particles:

If the parent 𝛼-emitter emits 𝛼-particles when it is in an excited state, then we get long range 𝛼-particles.  This is because the energy of excitation becomes available to 𝛼-particles as they reach the ground state of daughter nuclei.

The following figure illustrates the emission of long range 𝛼-particles from ₈₄Po²¹⁴.

𝛼ₒ - normal 𝛼-group corresponding to transition between ground states of  ²¹²Po and ²⁰⁸Pb.

𝛼₁, 𝛼₂ - these groups originate from transitions from excited states of parent ²¹²Po to daughter ²⁰⁸Pb directly.

Thus, nuclear spectroscopic studies of long range 𝛼-emitters provide information about nuclear energy levels of parent.

How to set trip point in overload relay of starter?

Well, most of the engineers, who use the motors find sometimes get quizzed that how they should find right trip point on the overload relay in the starter for the motor.

For this, one must find the current rating provided on the motor. This is usually the safe current which the winding of the motor can work with out failure. Usually the winding of motor can withstand high current also based on the time. If the motor curve is available it is easy to find out the required trip current based on the trip time.

or else, one has to consider the 1.5 times of the rated current of motor. When we set the trip point on the relay at the required current, the next step is to ensure the trip time. Trip time could be find out by using equivalent load.

The final step to ensure that you have used the proper relay is to perform rotor lock test
of the motor.        

What is rotor lock test?

The name itself gives and idea that it is locking of the shaft or rotor. One has to lock the rotor and switch on the motor.
This will lead to high current passage through winding. The relay should get tripped within prescribed time by manufacturer
to stop damage to the winding. This is the worse condition of failure and thus, suitable use of relay and trip set point, will always protect the motor.

Any queries are welcome.

Polar Dielectric in Uniform Electric Field

 i) There are permanent dipoles present in Polar Dielectric which are randomly aligned in such a way that there exists permanent dipole moment  Pp.

ii) When a dipole is present in an uniform electric field the dipole tries to align itself in the direction of electric field.

iii) Because of this all dipoles in polar dielectric are partially aligned in the direction of the field. This partial alignment is responsible for the induced dipole moment Pi.

Therefore the electric dipole moment is increasing.

P = Pp + Pi

iv) The electric dipole moment of a polar dielectric increases

      a) by increasing the applied electric field

      b) by decreasing the temperature 

Thermodynamics - important points to be noted for competitive exams

➔ Tephigram is the name of temperature entropy diagram

➔ PV graph in a adiabatic change is called Isentropic.

➔ Entropy of a system is an index of “unavailable energy”.

➔ When gas is expanded, work is done by the gas on surroundings.

➔ The size of “Kelvin degree” is equal to “Centigrade”.

➔ The efficiency of a carnot engine increases by raising the temperature of the source.

➔ Work done per cycle is given by the area enclosed in the indicator diagram.

➔ Conversion of heat energy into electrical energy can be made by “Thermocouple”.

➔ f(P,V,T) =0 exists for an equilibrium state and is called equation of state.

➔ The area of cycle of T-S diagram gives the “available thermal energy for useful work” in a reversible process.

➔ Uses of TS diagram: 

a) used in meteorology b) check efficiency of heat engine c) useful in predicting defects of performance of engine d) to obtain work value of fuel used.

➔ Change in entropy of universe due to free expansion is 
∆S = nR log e(Vf/Vi)

➔ Loss of available of energy = To.dS where ‘To’ is lowest available temperature in system.

➔ In order to maintain a body in an isothermal condition, heat has to be either supplied or withdrawn.

➔ When a gas expands adiabatically, the temperature decreases.

➔ When a gas is compressed, the temperature increases because work is done on the gas.

➔ The work done in an adiabatic change in a particular gas depends upon only change in temperature.

➔ In an adiabatic compression, the decrease in volume is associated with increase in temperature & increase in pressure.

➔ For an isothermal expansion of a perfect gas, the value of dP/P is equal to -dV/V. 

 

➔For an adiabatic expansion of a perfect gas, the value dP/P is equal to -𝛾dV/V.  

 

➔ A reversible process is always “quasi-static”, but every quasi-static process need not be a reversible process.

➔ For reversible cycle: ∆P = ∆V = ∆T = ∆U = ∆H

➔ dW = PdV is only applicable to reversible process. 

➔ In case of “irreversible processes”, dW is not equal to PdV; 

 

➔For free expansion, dW=0

➔For free expansion, dV=0, the work may be zero (in case of PV work)

 

➔ Work and heat are path functions.

➔ Work is not a thermodynamic property as it is not a state function and it is not a exact differential.

➔ Both thermodynamic and temperature scales use a single reference temperature i.e triple point of water.

➔ dW = PdV is only applicable to reversible process. 

➔ In case of “irreversible processes”, dW is not equal to PdV;