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The definition of beta $(\beta^-)$ decay is that an electron is produced.

For example,

$$\ce{^131_53I -> ^131_54Xe + e^-}$$

So iodine forms Xe and releases one electron with a -1 charge only and no weight.

My question is that if an electron is released, then there is still a proton left so shouldn't it still be iodine? Then you should gain an extra proton. I personally feel like something is missing here, Also about Positron Emission or Positron Decay.

Anyone know why or what?

Edit: The "duplicate" question picked by many does not fully answer the question as well as bon did here.


marked as duplicate by Klaus-Dieter Warzecha, LDC3, M.A.R., Freddy, Jannis Andreska Jan 11 '15 at 15:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


This is really a physics question but I will answer it anyway. I'm not sure how much subatomic physics you know so I will give two different versions. Simply put, a neutron decays to form a proton and electron. $$\ce{n -> p+ + e-}$$ This explains why the proton number increases by one to form Xenon.

More properly a down quark inside a neutron decays to form an up quark, an electron and an electron antineutrino as a result of weak nuclear force interactions. $$\ce{d -> u + e- + \overline{\nu}_{e}}$$

This causes the neutron ($\ce{udd}$) to turn into a proton ($\ce{uud}$) leading to the increase in atomic number.

This type of beta decay is know as $\beta^{-}$ decay. There is also $\beta^{+}$ decay which is a similar process where a proton becomes a neutron, positron and an electron neutrino: $$\ce{^23_12Mg -> ^23_11Na + e+ + \nu_{e}}$$ $$\ce{u -> d + e+ + \nu_{e}}$$

http://en.wikipedia.org/wiki/Beta_decay $%edit$

  • $\begingroup$ I'll just stick with the textbook and not ask any questions. Haha. Don't have time to comprehend when exam is in a couple of days. $\endgroup$ – Asker123 Jan 10 '15 at 21:35
  • $\begingroup$ Unless you have a particular need or desire to understand particle physics the first equation should allow you to work out whats going on $\endgroup$ – bon Jan 10 '15 at 21:36
  • $\begingroup$ Then why do they write the electron after the arrow, isn't it just with the Xe. I don't see the need to write the electron. Just say that Xe forms. $\endgroup$ – Asker123 Jan 10 '15 at 21:37
  • $\begingroup$ Could you give me an example of the $\ce{B^+}$decay. Like you explained with $\ce{B^-}$. $\endgroup$ – Asker123 Jan 10 '15 at 21:43
  • $\begingroup$ The electrons are ejected from the nucleus at very high energies, and therefore velocities, and so are not captured by the atom. This is why we usually write them seperately. $\endgroup$ – bon Jan 10 '15 at 21:43

When we talk about an unstable atom that will decay by emitting $\alpha$, $\beta$ or $\gamma$ radiation, the nucleus is the thing that's unstable, right?

As wikipedia puts it,

An unstable atomic nucleus with an excess of neutrons may undergo β− decay, where a neutron is converted into a proton, an electron and an electron antineutrino (the antiparticle of the neutrino)

You can forget the anti-matter part. Just remember than in $\beta$ decay there is a neutron from the nucleus that is being "devolved" into eminently a proton and an electron. The electron leaves the atom with high velocity and energy, but the proton remains. So, for each atom that a $\beta$ particle is released from, we have:

$$\ce{^A_zM ->^A_{z+1}M+}$$


Essentially, a neutron is converted to a proton, releasing an electron in the form of beta radiation. That is why the mass stays constant but the number of protons increases by 1.

  • $\begingroup$ I don't quite understand that, any clarification please. $\endgroup$ – Asker123 Jan 10 '15 at 21:17

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