Angular Momentum & Parity in Alpha decay
- 1. Angular Momentum & Parity in alpha Decay Presented By: D.Surat M.Sc. Physics
- 2. 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).
- 3. ❏ 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).
- 4. 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.
- 5. 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
- 6. 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.
- 7. ❏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 lf. ❏ Here many different states of 238Pu are populated. ❏ The alpha decays have different Q value and different intnsities.
- 9. ❏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
- 10. ❏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.
- 11. ❏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.
- 12. ❏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.
- 15. ❏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.
- 17. Reference ❏Introductory Nuclear Physics- Kenneth S. Krane ❏www.physics.isu.edu/radinf/alpha.htm ❏https://www.euronuclear.org/info/encyclopedia/alphaparticle.htm
- 18. Thank You !!