2
$\begingroup$

In beta decay neutrons convert into proton, in that process they'll release an electron (called beta particle) but mass still remains same
$$_zX^A \longrightarrow\; _{z+1}Y^{A}+ e^- $$ The question is where did that beta particle came out from no where like matter can't be created or destroyed?
Or am I misinterpreting something?

$\endgroup$
3
  • 4
    $\begingroup$ Matter can be created and destroyed - that's what radioactivity is. $\endgroup$ Jul 24, 2021 at 10:36
  • 1
    $\begingroup$ It should be noted that the neutron is more massive than the proton, so nucleus X is more massive than nucleus Y, though this difference is very small. $\endgroup$
    – Triatticus
    Jul 24, 2021 at 13:42
  • 1
    $\begingroup$ E=mc^2, so energy can be converted to mass (and vice versa) $\endgroup$
    – KingLogic
    Jul 25, 2021 at 3:51

1 Answer 1

7
$\begingroup$

In beta decay a neutron decays into a proton, an electron and an antineutrino. More precisely, one of the down quarks within the neutron changes into an up quark and emits a short-lived $W^-$ boson, which quickly decays into an electron and an antineutrino. The change of flavour by the quark turns the neutron into a proton.

The electron and the antineutrino do not have to "come" from anywhere. Transformations of one fundamental particle into another (or a set of others) can happen provided certain conservation conditions are met. In the case of beta decay we have

  • Conservation of charge - the charge of the neutron is $0$ and the net charge of the proton, the electron and the (uncharged) neutrino is $0$.
  • Conservation of energy - the neutron is more massive than the decay products, but the mass difference is accounted for by the kinetic energy of the decay products.
  • Conservation of momentum
  • Conservation of baryon number/quark number - the transformation leaves the number of quarks minus the number of antiquarks unchanged.
  • Conservation of lepton number - the lepton number of the neutron and the proton are $0$. The lepton number of the electron is $+1$, but this is balanced by the lepton number of the antineutrino, which is $-1$.
$\endgroup$
7
  • $\begingroup$ "Conservation of boson number - ..." You probably mean conservation of quark number (quarks are not bosons). $\endgroup$ Jul 24, 2021 at 11:41
  • $\begingroup$ @ThomasFritsch Yes indeed - fixed. $\endgroup$
    – gandalf61
    Jul 24, 2021 at 12:03
  • 1
    $\begingroup$ "the proton is more massive than the decay products" ... did you mean "neutron" instead of "proton"? $\endgroup$
    – Michael
    Jul 24, 2021 at 20:06
  • $\begingroup$ @Michael Yes I did. Fixed. $\endgroup$
    – gandalf61
    Jul 24, 2021 at 20:50
  • 1
    $\begingroup$ Thanks, @ThomasFritsch. The OP probably meant baryon number. $\endgroup$
    – J.G.
    Jul 24, 2021 at 21:08

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.