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The beta decay reaction is given by : ${ }_{Z}^{A} \mathrm{X} \rightarrow \underset{Z+1}{A} \mathrm{X}^{\prime}+\mathrm{e}^{-}+\overline{\nu}$. I would like to know the reason why cannot we write it like the below one as such for these reasons.

  • X has $Z e^-$ ,$Z p$ and $(A-Z)$ n
  • After decay happens one neutron converts to proton and a beta particle(antineutrino too) , so now the new atom X' has $(Z+1) p$ , $(Z)$ e^- (still) , $(A-Z-1)$ n . Hence the equation should look like this : ${ }_{Z}^{A} \mathrm{X} \rightarrow \underset{Z+1}{A} \mathrm{X}^{\prime+}+\mathrm{e}^{-}+\overline{\nu}$. but this is not what most textbooks/online write it as why thats the reason as such this equation satisfy the condition that charges are balances intially neutral finally too product is neutral (Thats why we also see the mass of X' is given for calculation of Q -value and not mass of X'+) . So why not write the reaction like this . Similar thing happens with positron decay too i would like to know the reason there too.
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Because, in most cases, you are concerned about what is happening with the nucleus.

$ ^{A}_{Z}X $ may even be an ion, but the beta decay equation will still be applicable. You really ignore what is happening in the electron shells.

You can always imagine, that a free electron will be drawn from the environment. If you ignore that part, then there is no reference to compare with, and thus there is no relative extra positive potential to add a $+$ sign to indicate an extra proton.

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  • $\begingroup$ That explains quite well , what do you mean by" no relative extra postive potential" Sean? And consider the Q value of the beta decay , in many books they consider nuclear mass of X' to be m(A Z+1 X') - (Z+1)m_e , shouldnt it be only (Zm_e) be subtracted as by the points i said above ? And are my all points are correct ? $\endgroup$
    – Orion_Pax
    Apr 13, 2022 at 4:31
  • $\begingroup$ And shouldnot in a nuclear charges in both sides be balanced too ? We can see its - in Right side but neutral in left $\endgroup$
    – Orion_Pax
    Apr 13, 2022 at 7:53
  • $\begingroup$ @Orion_Pax. Nuclear charges are balanced. $Z = Z+1+e^-$. We are however ignoring the neutral state of the atom. We call the neutral state 0 potential, because it is natural to assign 0 potential to zero charge. But we could as well move the 0 potential label anywhere else. We could call a nucleus with $Z$ atoms 0 potential. Because we ignore the neutral atom, there is no reference 0 potential. So you can't also say that $_{Z+1} A$ has 1 extra charge now. $\endgroup$
    – Sean
    Apr 13, 2022 at 8:41
  • $\begingroup$ Oh i see thats pretty nice reasoning , what about the Q value calculation Sean which i said above? $\endgroup$
    – Orion_Pax
    Apr 13, 2022 at 9:08

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