# Beta decay

1. Oct 24, 2014

### gxc9800

1. The problem statement, all variables and given/known data
from the physics fact in the photo, i know that the daughter nucleus is a positive ion....
but , referring to the part 2 , finding the mass loss in beta decay in terms of atomic mass , it says that A_(z+1) Y has extra one orbitting electron comapred to the parent nucleus.
this is contrary to the physics fact above.....
A_(z+1) Y is atom with proton number and number of electron = Z+1 or A_(z+1) Y is a positive nucleus ?

2. Relevant equations

3. The attempt at a solution

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2. Oct 24, 2014

### gxc9800

and i dont understand part 2 ) ,
why the mass defect is mx -(my-me) -me ?

the extra orbittin elc\ectron mass is already deducted in the last part -me am i right?
why the mass of elctron is deducted again in ( my-me )?

3. Oct 24, 2014

### collinsmark

The atomic masses of $^A_Z X$ and $^A_{Z+1} Y$ are things you can look up in a table of atomic masses. For example, let's take carbon twelve, $^{12}_6 C$. You can look that up and you'll find that the atomic mass is exactly 6 grams per mole (and divide by Avogadro's number to find the mass of a single atom). But that number assumes that the $^{12}_6 C$ atoms under consideration all have full shells of electrons (in this case 6). If you want to measure the atomic mass of a $^{12}_6 C ^+$ ion, with only 5 electrons, you'll need to subtract off the mass of an electron from the value found in the table.

The change in mass, $\Delta m$ is the atomic mass of the parent atom minus the atomic mass of everything involved after the decay.

So what is left after the decay? The atomic mass of $^A_{Z+1} Y$ with the electron subtracted off, and the atomic mass of the beta particle.

I'm not sure I understand. $m_e$ is the electron mass.

The byproducts of the beta decay consist of a positive ion and an electron. But the mass of $^A_{Z+1} Y$, which you can look up in a table, is not the mass of a positive ion. It's the mass of a neutral atom. So to find the mass of the positive ion, you must take the mass of $^A_{Z+1} Y$ and subtract off the mass of an electron.

4. Oct 24, 2014

### gxc9800

$A_{Z X}$ means $^{12}_6 C$ atom here? $^A_{Z+1} Y$ represent $^{12}_7 C ^+$ ion am i right? to find the mass of $^{12}_6 C$ atom, the mass os 1 electron is subtracted ... why you said that the mass is subtracted to find the mass of $^{12}_5 C ^+$ ion?

5. Oct 24, 2014

### collinsmark

I was just using carbon twelve as hypothetical example (and in retrospect, not such a good one).

$^A_Z X$ as expressed the problem can be any atom.

Well, if the parent atom was carbon, then the daughter atom would be a nitrogen ion, $^{12}_7 N^+$. Carbon isn't carbon if it has 7 protons. It's nitrogen. But let's pretend for the moment that the parent atom is some weird boron isotope with 7 neutrons, $^{12}_5 B$, then the daughter ion after a beta decay would be $^{12}_6 C^+$.

These are admittedly bad examples though since they are not typically atoms/isotopes one considers for beta decay.

A better example would be carbon fourteen.

$$^{14}_6 C \ \ \rightarrow \ \ ^{14}_7 N^+ + e^- + \bar \nu$$

Where the $\bar \nu$ is an anti-neutrino. But we're neglecting the mass of the anti-neutrio for this exercise because it its mass is negligible.

For example, here is a table of the masses of various atoms:

http://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl

It is assumed that the masses given are from atoms that are neutral (not ions). You won't easily find a table of masses of ionized atoms anywhere. So if you want to find the mass of a positive ion -- missing one and only one electron -- you can do that by looking up the mass of a neutral atom and subtracting off the mass of an electron.

6. Oct 24, 2014

### collinsmark

Just to follow up for the sake of clarity:

Continuing the carbon fourteed decay example, if you wanted to find the mass of the positive $^{14}_7 N^+$ ion, you could look up the atomic mass of the neutral $^{14}_7 N$ atom in the table, http://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl and then subtract off the mass of an electron.

But as the book points out, it's not necessary to do this when finding the $\Delta m$ because when considering the mass of the beta particle, the same as the mass of the missing electron, they cancel out.