In radioactive alpha decay, a helium atom is shown to be released.

However, I was told that only thing released is a helium nucleus. If so, then it should leave two of its electrons in the atom undergoing decay. Which would then mean that the element should possess an electric field around it since a helium nucleus (not a helium atom) is released.

Is that what happens?


You have been misled. An alpha particle decay is exactly that - the unstable nucleus emits an alpha particle, which is a He nucleus (2 protons and 2 neutrons). Radioactive decay is not an "atomic" process.

The emitted alpha particles (He nuclei) have a +2 positive charge. They leave behind a negatively charged daughter atom (a di-anion). The di-anion also won't be around long; it recoils from the alpha particle emission and will interact with its surroundings, passing on the spare electrons. It is partly this process that releases heat in radioactive materials.

The alpha particles do not go very far unless they are in a vacuum. In air, they travel typically a few cm before interacting with other atoms and picking up a couple of electrons to become He atoms.

| cite | improve this answer | |
  • $\begingroup$ so it does happen ...but only for a shortwhile. Right ?? $\endgroup$ – Vinayak Dec 27 '14 at 2:21
  • $\begingroup$ @Vinayak Yes, the atom that is left behind has two more electrons than protons. $\endgroup$ – Rob Jeffries Dec 27 '14 at 9:28

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.