# Defining Nuclear Reaction and Beta Decay

I use a school textbook that defines nuclear reaction in the following way. I will go through what does make sense to me, and then point out the thorns.

"When an atom changes into a different element, it is said to undergo a nuclear transmutation. In nuclear transmutations, electric charge is conserved".

So far, I think I am tracking. Atoms change into different elements (e.g. carbon-14 decays into nitrogen-14, a beta minus particle, and an antineutrino). Charge is conserved here as the charge of a neutron is zero (before decay), and the net charge of a proton and electron (after decay) is also zero. So no biggie. Nature has conserved charge despite the decay.

Something in the next line throws me off "In any nuclear reaction, including radioactive decay, atomic and mass numbers are conserved. Energy is released during these decays."

"mass numbers are conserved". Once again, no biggie. One neutron gets converted to a proton. Mass number remains the same since it is the total number of neutrons and protons.

"atomic ... numbers are conserved". Here's the thorn. In beta minus decay (i.e. decay that emits an electron), atoms increase in their atomic number. After a decay, there is one less neutron - but there is one more proton.

This leads me to think that beta minus decay doesn't result in the same amount of protons. Beta minus decay increases the protons using other things. There is no alpha particle, so it's not like there is another particle on the decayed side of the equation that I haven't considered. So atomic number is not conserved (but the mass number is).

Thus beta minus decay is not a nuclear reaction? But it happens to the nucleus! The nucleus is the source of the electron! Surely it is a nuclear reaction!

It may just be words - I am not even trying to understand the phenomena - but man am I invested in clarifying this definition.

• Does your textbook use the (weird) notation ${}^{\,\,0}_{-1} e$ for the electron emitted in a beta decay? Sometimes sources will use a convention where a beta particle is defined to have atomic number -1 in order to say that "$Z$ is conserved" in all reactions. Commented Apr 22 at 11:32
• It is extremely likely to be what Michael said. Especially since it is old enough to use "transmutation". Commented Apr 22 at 12:50
• Thank you @MichaelSeifert ! You guessed it $_{-1}^0 \beta$ is used to represent the electron produced in beta decay.
– HBP
Commented Apr 24 at 10:06
• @naturallyInconsistent Funnily enough, the textbook was written in 2016 ... or is that considered old in your book :P?
– HBP
Commented Apr 24 at 10:07

Fleshing out a comment as an answer:

In many sources, the beta particle emitted in a decay process is assigned $$Z = -1$$, and denoted $${}^{\,\,0}_{-1}e$$ or $${}^{\,\,0}_{-1}\beta$$ in decay processes. Similarly, inverse beta particles are assigned $$Z = +1$$.

This convention means that $$Z$$ is no longer an "atomic number" strictly speaking; electrons & positrons are not "atoms", after all. Instead, under this convention, $$Z$$ acts as a proxy for charge, and conservation of $$Z$$ implies conservation of charge.

• Thank you! The terminology can get so subtle at times ... don't know whether it's just the nature of English or people overcomplicate things or reality is just complicated ... It doesn't make sense to say that write electrons and whatnot like we would atoms, since they are just individual particles, and do not connect with others like in an atom/molecule. Anyway ... I ramble. Thanks again.
– HBP
Commented May 7 at 4:17

I am not a specialist, but with the notation you usually handle, what your book says does not seem correct: indeed, the atomic number changes in any of the beta processes. In particular, what is conserved is the baryonic number and the leptonic number.

• Thanks for your answer, thanks to Michael the confusion is resolved. I didn't read into the notation enough (i.e. $_{-1}^0 \beta$ the electron is said to have a negative atomic number!\$). Thanks for noting what is conserved (baryonic number and leptonic number), despite my utter unfamiliarity with those terms, which I will leave for an opportune time to learn about.
– HBP
Commented Apr 24 at 10:11