Beta Decay - Doesn't Add Up [duplicate]








**This question already has answers here**:
 
 




[Understanding Beta Decay](/questions/10453/understanding-beta-decay)

 (3 answers)
 

Closed 8 years ago.



The definition of [beta $(\beta^-)$ decay](https://en.wikipedia.org/wiki/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?




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




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> $%edit$


 5

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](http://en.wikipedia.org/wiki/Beta_particle) 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:


$$\ce{^A\_zM -> ^A\_{z+1}M+}$$


 3

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.


 0