Quick question on the mass condition for beta decay

[rwooduk]

For β^- we have:

$M(A,Z)>M(A,Z+1) + m_{e} - m_{e}$

An electron is removed from the atom and therefore we need to take that away from the M(A,Z+1) term

But for β⁺ we have been given:

$M(A,Z)>M(A,Z+1) + m_{e} + m_{e}$

What is this saying? A positron is emitted, therefore shouldn't we minus the mass of the positron from the atomic mass? which would give the same expression as for β^-? I'm confused as to why it is plus, it's probably something really simple but I can't figure it for some reason.

Thanks for any help/ideas

edit also a quick side question, is β⁺ decay the same as electron capture? As they give the same daughter nucleus.

 

[snorkack]

But for β⁺ we have been given:

$M(A,Z)>M(A,Z+1) + m_{e} + m_{e}$

What is this saying? A positron is emitted, therefore shouldn't we minus the mass of the positron from the atomic mass? which would give the same expression as for β^-? I'm confused as to why it is plus, it's probably something really simple but I can't figure it for some reason.

Thanks for any help/ideas

edit also a quick side question, is β⁺ decay the same as electron capture? As they give the same daughter nucleus.

Probably those who have given the equation are confused in explaining what the equation is for.
Positron decay and electron capture are not the same - daughter nucleus is the same, but other ingredients are not. They therefore have different mass conditions.

 

[rwooduk]

Probably those who have given the equation are confused in explaining what the equation is for.

So is the mass condition that we have been given for β⁺ decay incorrect?

Positron decay and electron capture are not the same - daughter nucleus is the same, but other ingredients are not. They therefore have different mass conditions.

Thanks! also i realized the spectrum is different, beta decay being broad and Auger giving discrete emission lines.

thanks for the reply

 

[mfb]

The beta+ equation is wrong in some way - if the left side is supposed to be the initial state (the ">" suggests this) then the charge at the right side is wrong. Fix that and the equation makes sense. Note that the number of electrons for the final atom (which is included in those mass values) is different for the two processes.

 

[rwooduk]

The beta+ equation is wrong in some way - if the left side is supposed to be the initial state (the ">" suggests this) then the charge at the right side is wrong. Fix that and the equation makes sense. Note that the number of electrons for the final atom (which is included in those mass values) is different for the two processes.

Thanks. Would you mind writing out the correct equation? I'm unsure what you mean by "the charge at the right side" when they are masses. thanks again for your help.

 

[mfb]

Z+1 in the brackets is the charge of the nucleus. If you emit a positive particle (a positron) the proton number has to decrease.
$$M(A,Z)>M(A,Z-1) + m_{e} + m_{e}$$The two more electron masses compared to beta- decay come from the additional positron and the fact that you suddenly have an additional electron that does not "belong to" the atom - so you have the new neutral atom plus one electron plus one positron, and the original atom has to have enough energy to produce all of them.

 

[rwooduk]

Z+1 in the brackets is the charge of the nucleus. If you emit a positive particle (a positron) the proton number has to decrease.
$$M(A,Z)>M(A,Z-1) + m_{e} + m_{e}$$The two more electron masses compared to beta- decay come from the additional positron and the fact that you suddenly have an additional electron that does not "belong to" the atom - so you have the new neutral atom plus one electron plus one positron, and the original atom has to have enough energy to produce all of them.

That's great thanks. Although I think I'm getting confused; Z is the number of protons, after decay it turns to a new element with Z-1 with an emitted positron, so then why are we adding the mass of an electron?

 

[mfb]

Let's consider an example:
$${}^{40}_{19}Ka \to {}^{40}_{18}Ar + e^+ + \nu_e$$
Potassium on the left side normally has 19 electrons, so we still have 19 electrons on the right side - but Argon just has 18 protons, so one electron is surplus. We can write the reaction as
$${}^{40}_{19}Ka \to {}^{40}_{18}Ar + e^- + e^+ + \nu_e$$
to get neutral atoms, because the mass numbers refer to those neutral atoms.

 

[rwooduk]

Let's consider an example:
$${}^{40}_{19}Ka \to {}^{40}_{18}Ar + e^+ + \nu_e$$
Potassium on the left side normally has 19 electrons, so we still have 19 electrons on the right side - but Argon just has 18 protons, so one electron is surplus. We can write the reaction as
$${}^{40}_{19}Ka \to {}^{40}_{18}Ar + e^- + e^+ + \nu_e$$
to get neutral atoms, because the mass numbers refer to those neutral atoms.

Of course, number of protons = number of electrons, so 1 electron is surplus. Sorry it's been 20 years since my GCSE chemistry, but still should know this lol

Many thanks!