5
$\begingroup$

When alpha decay happens the daughter nucleus is usually in ground state. When beta decay happens the daughter nucleus is usually in excited state which is then de-excited by emitting gamma photon.

Why is this?

$\endgroup$
  • 1
    $\begingroup$ There are many alpha decays that also emit gammas. Look through, say, the CRC table of isotope decays. $\endgroup$ – Jon Custer Jul 16 '16 at 20:40
  • $\begingroup$ @matori82 I'm no expert, but this LINK (below) en.wikipedia.org/wiki/Uranium-235#Natural_decay_chain indicates that U-235 decays to Th-231 by emitting an alpha-particle, and that Th-231 then decays to Pa-231 by emitting a beta-particle, and that the decay-process continues, by emitting either alpha or beta -particles, until it gets to Pb-207 [i.e., LEAD], which is stable ... So your statement that alpha-decay leads to ground-state is not true ... I know that, in the question, you said "usually": please give some examples re what kinds of decays you are asking about ..... $\endgroup$ – PERFESSER CREEK-WATER Jul 18 '16 at 23:07
4
$\begingroup$

The question uses the term "Usually" which is not a correct description , however the decay schemes can be understood by analzing the process in detail.

An alpha particle is identical to a helium nucleus, being made up of two protons and two neutrons bound together.

There are models in which a nucleus can be seen as cluster of alpha-particles; say Carbon -12 as composed of three alpha particles.

In the decay process it comes out from the nucleus of its parent atom, (invariably one of the heaviest elements) by quantum mechanical process of tunneling and is repelled further from it by electric force , as both the alpha particle and the nucleus are positively charged.

The process changes the original atom (its mass number decreasing by 4 and atomic number by 2) from which the alpha particle is emitted into a different element called daughter nucleus.

Sometimes one of these daughter nuclides will also be radioactive, usually decaying further by one of the other processes.

This tunneling through the barrier depends on the barrier potential defined by strong nuclear interaction and as the decay process is intended for stabilizing the nucleus to lowermost energy levels therefore many a time the daughter nucleus is found in the ground state but this energy transfer also leads to daughter being in an excited state and later reaching the ground state by emitting a beta particle or a gamma radiation.

Beta electron emission occurs by the transformation of one of the nucleus’s neutrons into a proton, an electron and an antineutrino.

Beta positron decays is a similar process, but involves a proton changing into a neutron, a positron and a neutrino.

The above process gets into motion for unbalanced nucleus where excess proton or neutron is found. The decay process is guided by weak interaction and the parity and angular momentum conservation/non-conservation are guiding principles which determine the transition to be allowed or forbidden.

The Q value of such reactions plays an important role and the presence of a free proton after its conversion and its spin relations with the associated electron plays a significant role as to the decay leading to a Fermi-transition or GT-allowed or a mixture of the two.

The beta decay process usually lands a daughter nucleus in an excited state from where it goes to ground or lower energy state by gamma transition.

The detail analysis of the transition is essential to find the final energy of the daughter nucleus. One can attribute it to the complexity of the beta decay process.

After a nucleus undergoes alpha or beta decay, it is often left in an excited state with excess energy and goes to stabilizing itself by gamma emission.

For a detailed analysis one can see: Chapter 8 Beta Decay (pdf) from a Nuclear Chemistry course by Loveland.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.