You will be required to write an investigative report on nuclear physics.
The idea of beta decay encompasses three kinds of nuclear transformations, electron (b-), decay, and positron(b?
decay and electron trapping.
There are about 900 beta radioactive isotopes.
Only 20 of these are known to be natural. The rest were synthesized artificially.
B– decay is the main cause of the majority of these isotopes.
In each case of b – decay, one electron is released.
It is theoretically possible to have double beta decay with two electrons(positrons), but it has not yet been observed experimentally (Detlafand, I?A.vorski?).
Energy Spectrum of Beta Decay
Beta decay’s energy spectrum is continuous and extends from E=0 through E=E0. The quantity E0 is known as the end-point energies of the beta spectrum.
E? E0/3 is E? E0/3 for heavy nuclei, and E?=0.25 – 0.45 MeV for natural b–radioactive element.
The E?0.5E0 energy spectrum for light nuclei is more symmetrical.
Half-lives of beta decays range from 2.5×10-2 second to four X1012 year, which is uncommensurably longer that the nuclear time (?10-21 to 10-22 second).
This means that beta decay can be explained by weak interaction.
This is a sign that beta decay is often accompanied with the emission of gamma-rays, which have a discrete spectrum of energy (Detlafand?A?vorski?)
(Figure 1 – Energy spectrum of the Beta decay electron
An electron antineutrino can be emitted with the electron in beta-decay, and an electronic neutrino along with a positron.
In comparison to the interaction of the nucleons and nucleons in a nucleus (nuclear interaction), the interaction between an electron neutrino, or antineutrino, with the nuclei is negligible.
In beta decay, the spins of electron (positron), electron antineutrino and neutrino are equal in magnitude and in opposite directions.
Therefore, the spin change of the nucleus equals zero.
The continuous spectrum of beta-death is caused by the differences in the energy distributions between the electron (positron), electron antineutrino, and neutrino (neutrino), the combined energy of these two particles being equal E0 (Detlafand?A?vorski).
The current concept states that the electron (positron), as well as the electron antineutrino, do not exist in atomic nuclear nuclei. They are formed at the instant of their emission from the nucleus due to weak interaction between the nucleons.
Because new particles are created in beta decay, non-relativistic quantum mechanics cannot be applied to this process. The problem is solved by quantum field theory.
The theory of beta-disintegration treats the production of an electronic and an electromagnetic neutrino (positron or electron neutrino) as the result if the interaction between the nucleon of a nucleus and the electron (positron), as well as neutrino and positron fields.
Apart from the production e- and/or u?e particles,
Other than the production of e?e and u??e (or e?
The weak interaction constant g, which is the coupling constant of electron-positron fields and nucleons, determines the intensity of this interaction. g1.4 X 1049 erg.cm3.
containing: a wave function of the nucleon in the initial state i; wave functions of the nucleon, electron (positron) and electron neutrino (anti-neutrino) in the final state k; interaction energy corresponding to the transition i -k; and, finally, a quantity determining the density of the number of final states of the system.
Selection rules for beta decay determine a much higher probability of allowed transitions than a low probability so-called forbidden beta transformations (Detlafand?A?vorski?)
Analyzing the energy spectrum N (E) is essential in the study and analysis of beta decay. N is the number or positrons emitted.
The N(E) distribution curves of beta spectra allow them to be classified into Fermi-Spectra and Forbidden spectra.
Permitted spectra are distinguished by the degree of their forbiddenness.
Allowable beta spectra are those whose mass is equal to zero.
where p andE = momentum, energy and electron momentum in units of mec2 or mec2
me= rest mass for the electron
E0= The end-point, or maximum energy, of the electrons and positrons of the beta spectrum.
F(Z.E), which takes into consideration the influence of nuclear fields on the curve of N(E), is the function.
N(E) includes a factor for forbidden beta spectra. It depends on E0,E and the degree or forbiddenness.
To determine whether a beta spectrum is Fermi, or forbidden, a Fermi–Curie plot must be created.
Nexp(E), the observed curve of beta spectrum, is where you will find it.
K(E), for Fermi’s beta spectrum, is a straightline which intersects E=E0’s abscissas axis.
If K(E) is not a straight line, it means that the beta spectrum given is forbidden.
The decay constant l is for beta decay
l = C= CF (ZE0)
The theory of beta decay includes the factor C.
Where g=weak interactions constant (coupling consta)
Since l=ln2T1/2, where the half-life is T1/2, then
The reduced half life of the product FZ,E0 T1/2=T0.5red can be described.
It all depends on the nature and interaction between the nucleons from the nucleus, the electron-neutrino force, and the nucleons.
to be determined.
The following types of beta-death can be classified according to their T0.5 Red values:
values are near to the maximum values;
LogT5red5 holders are allowed to make transitions.
LogT1/2red9-13 and 18 represent forbidden transitions of the 1st to 3rd order of forbiddenness.
These last cases represent a sharp drop in probability of beta decay due to large fluctuations in the angular momentum (and, more often, the change in its states’ parity).
The decay constant, lE, for an allowed electron catch is
where FE (e0,Z =2P ()3 [e0+1 -]2
This formula does not account for the relativistic effects, which are apparent when E0 approaches 0.511MeV rest energy.
The quantum system can be described as being even if the wave function that corresponds to it is not inverted in sign when all of the coordinates of the particles within the system are changed (in an inversion); otherwise it would be considered odd.
Parity P, where P = 1, can be used as a sign to describe the conservation of wave function sign after a space inversion.
Parity is defined as the opposite sign of the coordinates of a wave function. This is because the sign is reversed when the signs are changed.
The concept of parity in state and wave function is related to the space symmetry, i.e.
to the equivalence of space in the right- and left hand directions, upward or downward, respectively.
Any system of particles may be found in a condition with a definite paraity if it has a constant number of particles or is altered by an even number.
According to the Schrodinger equation, if the energy of any particle (or group of particles) is preserved constant, then its Parity Conservation Law applies.
A system of particles should have the same wave function (nucleus prior to decay, nucleus, beta particle, and (anti-neutrino), neutrino after decay).
It has been shown that decay can occur in weak interactions (e.g., those that cause beta decay)
A state that is described as an “even function” can also be called an “odd state”.
This phenomenon was first found in the beta-death of K mesons.
This basically refers to the conversion of a K particle into two or more pi mesons.
The violation of parity conservation can be seen in beta decay in the asymmetry in the spatial directions of electrons emitted. Less electrons are emitted towards the spin direction of nuclei than the other direction.
This asymmetry reveals that there is an inverse relationship between the direction and spin of the particle, as well as the direction of its space motion.
The spins of an antineutrino or neutrino should always be aligned parallel to their direction of motion. In other words, the antineutrino will be oriented along its direction of motion while the neutrino will be in the opposite direction (longitudinalpolarization).
If spin is compared to rotation then the motions for the antineutrino are the motions for the neutrino and right-hand spin.
The distinction is therefore made between left-hand helicity and right-hand helicity for the antineutrino.
Research results based on radiation from radioactive substances passing through electric and magnetic fields were the best.
The experiment showed that the fields were indeflected by the gamma-rays.
However, the beta and alpha rays deflected in opposing directions.
This proves that the gamma and alpha rays are not charged but they are charged.
The bullet’s penetration rate will vary depending on the material used to make the bullet and the material used to target it.
For example, an airgun pellet will only stop at a few millimetres of wooden material, while a high powered rifle bullet will pass through many thousands of millimeters steel.
In the case of ionizing radio, it may be similar.
Different ionizing radioactive radiations have different penetrating abilities (a- to b- and g-radiation).
This is an illustration of the radiation penetration from radioactive sources.
Radioactive alpha particles have a range of 10 cm or less in air.
Condensed matter medium (e.g.
The range of condensed matter mediums (e.g. water and tissue) is even smaller.
It is clear that a -particle will not penetrate clothing.
You can sit on any radioactive source without causing any damage to yourself if only the -particles were emitted.
However, if the source is within our bodies, all of the energy is deposited inside and can cause a high radiation dose.
Aleksandr Litvinenko (2006) with Po-210.
Po-210 emits an Alpha particle with energy 5.3 MeV
Beta-particles are a type of particle with a range of about 5 mm in soft tissues.
Most b-particles have a lower energy than 1 MeV.
Therefore, all b particles that come from environmental sources are stopped by our clothing
Gamma-radiation is able to penetrate tissue and concrete in our bodies.
The g-rays of Cs-137 are 50% and have an energy level of 0.662 MeV. It can penetrate water layers of approximately 9 cm. (known as the half-value).
It doesn’t matter if the source is outside or inside your body, g-radiation measurements are easy to make.
Furthermore, medical diagnostics can also be performed using g-radiation emitting isotopes.
The binding energy of the nucleus determines which decay process is possible, if any.
In the 60Co27 nuclear decay into the 60Ni28 nuclear nucleus, an b-particle is released, i.e.
The three products release an energy of 1.56 MeV as kinetic energies.
The b?- particle emission from 22Na11 decay is illustrated in this example:
22Na11-22Ne10 + E?+ue -0.55 MeV
A further example of electron trap is the transformation of the 37Ar18 nuclear nucleus into a 37Cl17 nucleus by the atomic electron that has been absorbed.
This reaction produces 0.0186 MeV energy.
A nucleus located above the area of stability, i.e.
A region of neuron excess
Reduce the number neurons within it to make it more stable.
This is how a nucleus that has been disintegrated in this way reduces its N/Z.
This is accomplished by the b-decay. A neuron within the nucleus transforms into an electron, creating and emitting an em and a neutrino simultaneously.
Similar to the above, a nucleus that lies below the stability area becomes stable by increasing its ratio N/Z.
A nucleus below the stability region can be made stable by increasing its N/Z.
A proton in a nucleus transforms into a neutron during the b-decay. The process simultaneously creates an antineutrino and a protons.
The electron capture process involves the combination of an atomic orbital electron and a proton from the nucleus. This results in the proton becoming a neutron and neutrino. Mani and Mehta (1990).
To understand the energetics of beta-decay, we use the atomic mass to create the mass-energy equations.
The neutrino mass is not shown in these equations because it is non-existent.
Q is the energy available to decay.
It is notable that the electron mass doesn’t appear on (1)’s right-hand side.
Because an electron is necessary for the atom AMZ+1’s to become neutral, this is why he is so.
There are two electron masses on the right side of the equation for b?- decay.
These two electron masses are due to the positron emitted by the nucleus and the second due to an extra electron (in AMZ-1) that is released by the atom to make it neutral.
Electron capture eq(3) is where an orbital electron of the nucleus is captured and the atom AMZ-1 is created. This is indicated by the superscriptasterisk (Mani, Mehta and 1990).
Let’s take a nucleus with Z protons, N neutrons.
It will be difficult to separate the nucleus into its individual protons or neutrons. These particles are held together with attractive forces.
The nucleus needs energy to be separated into its constituents. This allows each component to move to infinity.
This energy is known as the binding energy.
The binding energy of the nucleus is the energy required to separate the particles.
The Einstein mass energy relationship states that their combined mass in the coalesced condition must be lower than the sum of their masses if they are far from each other.
Dmc2 denotes the loss of mass, which is also the binding energy.
BE = Dmc2.
If M,Mn,and Mp are respective masses of a nucleus or neutron (AXZ), then the mass defect Dm will be.
Dm= ZMp + NMn – M
The binding energy of a nucleus is also known as
BE =(NMn +ZMp – m)c2.
The energy of the state at rest of infinite nucleon separation is assumed to be zero.
It is much easier to work in the mass of anatom than the nucleus.
We add and subtract the Z electron mass from the equation above to do this.
BE = (NMn + ZMp+ ZMe – M – ZMe).c2.
ZMp+ ZMe= Z (Mp+ Me).
Because the binding energy is much lower than (Mp+Me)c2, the quantity Mp+ Me is only the mass one atom of hydrogen (MH).
In the same way, the mass of the AXZ (MA) atom (M(AXZ?)+ ZMe is)).
Accordingly, the binding energy can be described in terms the the atomic masses.
The nucleus can have both a mass defect and a binding energy.
A binding energy is a combination of an electron in anatom, an electron in a molecule and a molecule within a crystal. It can be found in any atom or molecule. But it is rarely so small that one can ignore its mass equivalent.
Because of the strength of the nuclear forces, the binding energy of a atom is much smaller than its total mass.
The average binding energy per nucleon is B.
M is the mass the atom.
Which is the value of A for all known nuclei?
If we plot B for all A values for known nuclei, then the curve is as follows:
( Figure 2: B
it plotted for all the values of A for known nuclei
Strontium -90 A Particular Element
Strontium 90 is a rare element.
When Plutonium or Uranium undergo a nuclear fission reactions, Strontium becomes a byproduct.
A large amount of 90Sr was actually produced in the nuclear weapons tests of 1950 and 1960.
Natural Strontium does not have radio-active property, but its isotopes do.
They undergo beta-decay.
It has a decay energie of 0.546 MeV. It is found among an anti-neutrino and an electron.
The basic equation is
It has a half-life approximately 28.8 Years, meaning that it takes 28.8 Years to decay to half of its initial level.
Strontium can be converted to Yttrium90 with a half life of 64 years.
90Y is then converted to Zirconium (Zr), a stable substance. (“Strontium 90 Half-Life, Properties Beta-Decay and Uses
The beta particles radiated from Ytrium have high energy, while those of Zirconium are moderate in energy (“Radiation Protection
Strontium is a soft, shiny metal with a similar colour to lead.
It does not emit gamma decay but only beta decay.
Its chemical properties are comparable to Calcium’s and have the potential to precipitate in bones and teeth of animals (US EPA 2018, 2018).
It is very dangerous for our health and can be inhaled together with water.
It is easy to be absorbed by bones if it is accidentally consumed.
Because of its long half-life, it continues to harm our health.
It causes Leukemia and cancers of bones.
A., and I?A.vorski?
Handbook of Physics. Translated by Nicholas Weinstein, 3rd Ed.
London: Mir Publications.p.
New Delhi, Affiliated East West Publishing PVT Ltd.ISBN10: 8185095736 / ISBN13: 9788185095738
Half-life, Properties, Beta Decay, Uses of Strontium-90
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