What is the Charge Change in Beta Decay of Sodium to Magnesium?

AI Thread Summary
In beta decay, sodium-24 decays into magnesium-24, resulting in a charge change due to the conversion of a neutron into a proton, which increases the charge by +1. The emitted beta particle does not affect the charge of the magnesium nuclide itself, as it is a product of the decay process. While the magnesium atom may temporarily exist as a +1 ion, the notation {}^{24}_{12}Mg refers specifically to the nucleus, not the overall atom. The discussion also notes that the emitted electron typically has enough energy to escape the atom, but this does not influence the decay process. Overall, the charge change is understood within the context of nuclear decay rather than atomic behavior.
Matt-235
Messages
17
Reaction score
2
Okay, so one of my hall mates across the way asked this question about his Chemistry, and then I realized that I wasn't too sure why it was myself, so I figured I'd pass this on to you.

Take the decay equation

<br /> {}^{24}_{11}Na \rightarrow {}^{24}_{12}Mg + {}^0_{-1}\beta<br />

His question pertained to the overall charge of the nuclide in question. He mentioned that since a neutron is converting to a proton, there should be a net charge of +1 now on the Magnesium (since the {}^{0}_{-1}\beta comes from the conversion of a netron into the proton, that doesn't get factored in). Is it just that the charge is irrelevant to the decay that it's not bothered to be included?
 
Engineering news on Phys.org
A minor point is that the right side should also have an antineutrino. Physically speaking, the newly formed magnesium atom is likely to be a +1 ion for a while, the amount of time it takes to neutralize depending on its environment. But I think when they write {}^{24}_{12}Mg, the context is understood to be the nucleus itself, not the atom as a whole.
 
Okay, that's what I figured, I was just wanting to make sure that there wasn't something I was missing. Thanks.
 
To anticipate a possible extension, the electron emitted by the nucleus usually has a high enough energy to escape the atom, although it's also possible for the neutrino to carry the lion's share of the energy released in the decay. But it's not really relevant; what happens in the electron shells has no impact on the decay process.
 
Hello everyone, I am currently working on a burnup calculation for a fuel assembly with repeated geometric structures using MCNP6. I have defined two materials (Material 1 and Material 2) which are actually the same material but located in different positions. However, after running the calculation with the BURN card, I am encountering an issue where all burnup information(power fraction(Initial input is 1,but output file is 0), burnup, mass, etc.) for Material 2 is zero, while Material 1...
Back
Top