What if the nucleus has a magnetic moment, and also the electron shell has one? I suspect, in this case, the orientation of the nucleus could be "fixed" by the electron shell.

Maybe a mono-crystal of such a material would have an anisotropic radioactivity. Thus, its radioactivity wouldn't be the same in all directions.

I think, relative many requirements should be fulfilled:

  • The nucleus must be radioactive.
  • It must have a magnetic moment.
  • The electron shell must have also a magnetic moment.
  • The orientations of the atoms in the crystal structure should be unidirectional.

Is it possible? Does any such crystal already exist? Maybe the crystal of a radioactive isotope of some ferromagnetic element could fulfill all of these?

If not, why not?

  • 1
    $\begingroup$ Your best guess is probably a neutron star. $\endgroup$ – CuriousOne Jun 18 '16 at 9:06
  • $\begingroup$ @CuriousOne :-) Well, currently I am thinking on laboratory-sized mono-crystalls. :-) $\endgroup$ – user259412 Jun 18 '16 at 9:26

The idea that you propose is possible. In fact, there is a very famous example of your setting: the Wu experiment.

Chien-Shiung Wu at the US National Bureau of Standards prepared a thin surface of ${}^{60}Co$. This isotope decays by beta decay, producing one electron and one antineutrino. Due to the small magnetic moment of the nuclei, they had to cool the surface to 0.003 K, and then the sample was magnetized in an uniform magnetic field.

The result was quite amazing: the emitted electrons showed a preference to be in the direction of the nuclear magnetic moment, what proves that parity symmetry is violated. But that's another story...

More info: Wu experiment (wikipedia)

  • $\begingroup$ Thanks, ok. But the cause of the anisotropy here was the P-violation of the weak interaction. Couldn't be the cause of the anisotropy a magnetic interaction between the nucleus and electron shell? $\endgroup$ – user259412 Jun 18 '16 at 15:32
  • $\begingroup$ I asked for the possibility of the anisotropic radioactivity, where the source to the anisotropy is the internal structure of the nucleus and not the P-violation of the weak interaction. But I didn't mention it enough clearly in my question, so thank you very much your answer. $\endgroup$ – user259412 Jul 25 '16 at 22:17

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.