Alpha Decay & 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 !!