Question

`beta^-` decay problem?

In `beta^-` decay a neutron in a nucleus is transformed into a proton, an electron, and an antineutrino:

`n rarr p + e^(-) + barv`

(a) The physical quantity not conserved in `beta^-` decay is____

(b) The particle denoted by the question mark in the reaction

`N_7 ^(13)" rarr` `C _6 ^(13)" +e^(-) + ?`
is____.

Note that the superscript and subscript in `C _6 ^(13)"`, for instance, are respectively the mass number A and atomic number Z of the carbon nucleus.

Answer 1

As shown in the question, in `beta^-` decay, a neutron splits into a proton (which stays in the nucleus) and emits an electron and antineutrino:

`""_(0)^(1) n -> ""_(1)^(1) p + ""_(-1)^(0) e + barupsilon`

where `""_(0)^(1)n` is a neutron, `""_(1)^(1) p` is a proton, and `""_(-1)^(0)e` is an electron.

Part (a) shows that the atomic number is not conserved in `beta^-` decay, otherwise known as `beta` decay. Actually, it increases `Z` by `1` and keeps `A` the same.

In an explicit example, the question presents:

`""_(7)^(13) "N" -> ""_(6)^(13) "C" + ""_(-1)^(0) e + ?`

There may be a typo in the problem... clearly, `6 - 1 ne 7`. I think it should be...

`""_(6)^(13) "C" -> ""_(7)^(13) "N" + ""_(-1)^(0) e + ?`

In THAT case, it would be an antineutrino. It IS true that `7 - 1 = 6`...