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While reading my text I came across two statements as follows :

  1. Speed of alpha particle is a characteristic speed as it depends on nature of nucleus.

  2. Beta particle come out from same nucleus with different speeds so it is not a characteristic speed.

Now I am unable to understand why alpha particle speed depends on nucleus and why not then beta also ?

Are these just observation or there is something theoritically to say about these observations ?

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    $\begingroup$ Gawh! What horrible and mostly useless phrasing. They could make the same point so much clearer just by exhibiting the energy spectra. See physics.stackexchange.com/questions/123823/…? and physics.stackexchange.com/questions/189664/…?. $\endgroup$ – dmckee Mar 13 '17 at 5:13
  • $\begingroup$ @dmckee I am not able to get the energy spectra point , I read there about it $\endgroup$ – Physicsapproval Mar 13 '17 at 5:18
  • $\begingroup$ The accepted answer in the first one is pretty explicit. $\endgroup$ – dmckee Mar 13 '17 at 5:19
  • $\begingroup$ I don't know about uncoupled equations and what was the term $\sqrt {(p_e)^2 + (m_{e^-})^2} $ $\endgroup$ – Physicsapproval Mar 13 '17 at 5:32
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When a nucleus decays the reaction is characterised by the release of a fixed amount of energy called the Q-value of the reaction.

This diagram shows what happens when AM-241 decays to Np-237 with the emission of an alpha particle.

Am241->Np237 decay diagram

The energy levels in both nuclei are well defined and so the energies of the alpha particles are well defined.

$\rm energy_{\rm decay} =energy_{excited \,daughter} + energy_{alpha}$

So in the example shown the energies of the emitted alpha particles will be 5.48, 5.54, 5.58, 5.61 and 5.64 Mev.
The excited daughter nucleus then gets rid of the surplus energy with the emission of a gamma.

However for beta decay the quantum jumps as characterised by those shown below for alpha particles are accompanied by the emission of two particles which together carry away a fixed amount of kinetic energy.
The difference is that the sum of the kinetic energy (beta) and the kinetic energy (anti neutrino) is fixed but the energies of the emitted particles is not.
This means that you can have a range of beta energies and a corresponding range of anti neutrino energies.

decay sheme of Co60

In this example there would be a range of beta energies from zero (approx) to 0.31 MeV for one decay mode and from zero to 1.48 Mev for the other decay mode. The rest of the energy being taken away by the antineutrino.

$\rm energy_{\rm decay} =energy_{excited \,daughter} +energy_{\rm beta}+ energy_{\rm antineutrino}$

Again the excited daughter nucleus gets rid of the surplus energy via gamma emission

So it is the maximum energy of the beta particles which is the characteristic of the decay.

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  • $\begingroup$ why the image shown shows gamma particle then? as you say "range of beta energies and a corresponding range of anti neutrino energies." is it beta particle in image? $\endgroup$ – Physicsapproval Mar 13 '17 at 6:30
  • $\begingroup$ @Physicsapproval I have added a beta energy diagram to my answer. $\endgroup$ – Farcher Mar 13 '17 at 6:38
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    $\begingroup$ Maybe it would be clearer if you stated that it is the kinematic difference between two body decays and three body decays, the beta decay of the neutron as prime example. $\endgroup$ – anna v Mar 13 '17 at 6:40
  • $\begingroup$ @Farcher So all what causes different Beta particle energies is due to the neutrino and antineutrinos can have a range of momnentum and energy ? $\endgroup$ – Physicsapproval Mar 13 '17 at 6:52
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    $\begingroup$ @Physicsapproval Yes, so if the beta has an energy of 0.22 MeV the antineutrino will have an energy of 0.09 MeV if the total available energy was 0.31 MeV. $\endgroup$ – Farcher Mar 13 '17 at 7:06
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Both of them depend on the nucleus: when it decays, it releases some energy. The question is how is that energy distributed among the decay products.

The basic difference arises because $\alpha$ decay is a two-body decay (initial nucleus $\rightarrow$ final nucleus + $\alpha$), whereas $\beta$ decay is a three-body decay (initial nucleus $\rightarrow$ final nucleus + electron + (anti) neutrino). Conservation of momentum and energy then nails down the momentum of the $\alpha$ in the first case, but not that of the electron in the second case since the neutrino can account for any "missing" momentum (in fact, that's why Pauli proposed the neutrino in the first place: to account for the missing momentum).

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It determines the ionization in a medium. A slower velocity leads to higher ionization per unit path. As opposed to $\beta$, the path of $\alpha$ is more of a straight line. When a nucleus emits an alpha particle then in this process two particles are involved. The alpha particle and the residual nucleus. Now both these particle share the disintegration energy. They do it with the alpha taking a definite energy and the residual nucleus the remaining based on momentum and energy conservation. All alpha particles that are involved in this particular disintegration process take the same energy and therefore the speed. So we say the speed is a characteristic property. It determines beside the ionization the energy levels involved in the emission of alpha particle. This is not so for the beta particle where the disintegration energy is shared by three particles. The additional particle is the neutrino. Different betas from the same disintegration have different energies. Therefore given the energy of beta particle one cannot know for sure the energy levels involved in the disintegration process.

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  • $\begingroup$ Sorry but could you edit or clarify your post and I'm not sure how it addresses the question. $\endgroup$ – ZeroTheHero Mar 13 '17 at 22:45
  • $\begingroup$ I hope it clarifies. $\endgroup$ – SAKhan Mar 14 '17 at 6:00
  • $\begingroup$ Initially my answer had the first two lines. I later added other lines to clarify. I have addressed the question as to why speed of alpha is a characteristic property. The answer is in essence the same as that given by Nick. $\endgroup$ – SAKhan Mar 18 '17 at 16:14

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