Angular Momentum & Parity in alpha

DecayPresented By:

D.SuratM.Sc. Physics

What is Alpha Decay ?❏Alpha decay is a radioactive process in which a particle with two

neutrons and two protons is ejected from the nucleus of a radioactive atom. The particle is identical to the nucleus of a helium atom.

❏Alpha decay only occurs in very heavy elements such as uranium, thorium and radium. The nuclei of these atoms are very “neutron rich” (i.e. have a lot more neutrons in their nucleus than they do protons) which makes emission of the alpha particle possible.

❏After an atom ejects an alpha particle, a new parent atom is formed which has two less neutrons and two less protons. Thus, when uranium-238 (which has a Z of 92) decays by alpha emission, thorium-234 is created (which has a Z of 90).

❏ It consists of two neutrons and two protons, and is thus identical to the nucleus of a helium atom. The rest mass of the alpha particle amounts to 6.64424·10-27 kg, or 3.7273·109 eV.

❏ Alpha radiation is the radiation with the lowest penetration potential of the three radiation types (alpha, beta, gamma radiation).

History❏Alpha particles were first described in the investigations of

radioactivity by Ernest Rutherford in 1899, and by 1907 they were identified as He2+ ions.

❏By 1928, George Gamow had solved the theory of alpha decay via tunneling.

❏The alpha particle is trapped in a potential well by the nucleus. ❏Classically, it is forbidden to escape, but according to the (then)

newly discovered principles of quantum mechanics, it has a tiny (but non-zero) probability of "tunneling" through the barrier and appearing on the other side to escape the nucleus.

Cont’d❏Gamow solved a model potential for the nucleus and derived, from

first principles, a relationship between the half-life of the decay, and the energy of the emission, which had been previously discovered empirically, and was known as the Geiger–Nuttall law

Angular Momentum and Parity❏Let us consider alpha particle undergoing transition from initial

nuclear state of angular momentum Ii to final state If.

❏The angular momentum of alpha particle can range between Ii+If and |Ii-If|.

❏The nucleus 4He consists of two protons and two neutrons.❏ All in 1s states and with their spin coupled pairwise to zero.

❏ The spin of the alpha particle is therefore zero, and the total angular momentum carried by an alpha particle in a decay process is purely orbital in character.

❏We will designate this by lalpha.

❏The alpha particle wave function is then represented by a Ylm with l=lalpha.

❏Parity associated with alpha emission is (-1)Lalpha .

❏we have parity selection rule,indicating which transitions are permitted and which are absolutely forbidden by conservation of parity.

❏If the initial and final parities are same, the lalpha must be even; if the parities are different, then lalpha must be odd.

❏We know givin initial state can populatr many different final states in he daughter nucleus, this propertie is called fine structure of alpha decay.

❏Figure below shows the alpha decay of 242Cm

❏ Initial state is zeroand thus angular momentum of alpha particle lalpha is equal to the angular momentum of the final nuclear state l f.

❏ Here many different states of 238Pu are populated.

❏ The alpha decays have different Q value and different intnsities.

❏ Intensity depends on the wave function of the initial and final states, and also depends on the angular momentum lalpha.

❏Centrifugal potential l(l+1)h2/2mr2 must be included in spherical coordinates, which is always positive.❏ This has the effect of rising the potential energy for a<r<b and thus

increasing the thickness of the barrier which must be penetrated.

❏Lets take an example, consider the 0+, 2+, 4+, 6+ and 8+ states of the ground state rotational band. The decay to the 2+ state has less intensity than the decay to the ground state for two resons-❏ The centrifugal potential raises the barrier by amount 0.5MeV. And,

❏ The excitation energy lowers Q by 0.044Mev.

❏The decay intensity continues to decrease for these same reasons as we go up thwe band to the 8+ state.

❏If weuse ourprevios theory for the decay rates, taking into account the increasing effective B and decreasing Q, we obtain the following estimates for the relative decay branches: 0+,76%; 2+,23%; 4+,1.5%, 6+,0.077%; 8+,8.4*10-5.

❏These results are not in exact agreement with the observed deacy intensities, but they do give us rough idea of the origin of the decrease in intensity.

❏Once we go above the ground state band, the alpha intensities become very small, of theorder of 10-6 of the total decay intensities.

❏This situation results from the poor match of initial and final wave function.

❏There are some states for which for which there is no observed decay intensities at all.

❏Shown in fig above.

❏Alpha decay to these states is absotulety forbidden by selection rule.

❏For example a 0->3 decay must have lalpha=3.

❏Which must give a change in parity between initial and final states.

❏Thus 0+->3- is possible, but not 0- ->3-.

❏Similarly, 0->2 and 0->4 decays can not change the parity, and so 0+->2- and 0+->4- are not permitted.

❏When neither initial nor final states have spin 0, the situation is not so simple and there are no absolutely forbidden decays.

❏For example , the decay 2-->2+ must have odd lalpha and the angular momentum coupling rules require 0<=lalpha<=4.

❏ Thus it ispossible to havethisdecay with lalpha=1 or 3.

❏ Whether lalpha is 1 or 3 ?

❏ From previous discussion we may say that lalpha =1 intensity is roughly an order of magnitude grater than that of lalpha,=3 intensity.

❏Measuring only the energy or the intensity of the decay gives us no information about how the total decay intensity is divided amoung the possible values of lalpha.

❏To make the determination of the relative contributions of the different l values, it is necessary to measure the angular distribution of the alpha particles.

❏The emission of an l=1 alpha particle is governed by a Yl(theta,phi), while an l=3 alpha decay is emitted with the distribution according by a Y3(theta,phi).

❏If we determine special distribution of these decays. We could in principle determine the relative amounts of different l values.

❏As an example of such an experiment. We consider the decay of 253Es to states of the ground state rotational band of 249Bk.❏ The possible l values are inicated in figure below and the result of

measuring the alpha particle angular distributions help us to determine the relative contribution of the different values of lalpha.

❏Since many alpha emitter are deformed, these angularmomentums can also help us to answer another question: if we assume a stable prolate nucleus. Will more alphas’s be emitted from poles or from the equator ?

❏Figure above suggests a possible answer to htis question:❏ At larger radius of poles the alpha particle feels a weeker coulomb

potential and must therefore pnetrate a thinner and lower barrier

❏ We therefore expect that polar emission out tobe more likely than equitorial emission.

❏Figure below shows the angular distribution of alpha emission relative to the symmetry axis.

❏You may see tht emission from the pole is 3-4 times more probable than emission from the equator.exactly as we expect on the basis of the potential.

Reference ❏Introductory Nuclear Physics- Kenneth S. Krane

❏www.physics.isu.edu/radinf/alpha.htm

❏https://www.euronuclear.org/info/encyclopedia/alphaparticle.htm

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