Mass Loss in Beta Decay: A_(z+1) Y vs Physics Fact

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Homework Statement


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 ?

Homework Equations

The Attempt at a Solution

 
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and i don't 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 )?
 
gxc9800 said:

Homework Statement


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 ?

The atomic masses of [itex]^A_Z X[/itex] and [itex]^A_{Z+1} Y[/itex] are things you can look up in a table of atomic masses. For example, let's take carbon twelve, [itex]^{12}_6 C[/itex]. 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 [itex]^{12}_6 C[/itex] atoms under consideration all have full shells of electrons (in this case 6). If you want to measure the atomic mass of a [itex]^{12}_6 C ^+[/itex] ion, with only 5 electrons, you'll need to subtract off the mass of an electron from the value found in the table.

The Attempt at a Solution

gxc9800 said:
and i don't understand part 2 ) ,
why the mass defect is mx -(my-me) -me ?

The change in mass, [itex]\Delta m[/itex] 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 [itex]^A_{Z+1} Y[/itex] with the electron subtracted off, and the atomic mass of the beta particle.

the extra orbittin elc\ectron mass is already deducted in the last part -me am i right?
I'm not sure I understand. [itex]m_e[/itex] is the electron mass.

why the mass of elctron is deducted again in ( my-me )?


The byproducts of the beta decay consist of a positive ion and an electron. But the mass of [itex]^A_{Z+1} Y[/itex], 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 [itex]^A_{Z+1} Y[/itex] and subtract off the mass of an electron.
 
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collinsmark said:
The atomic masses of [itex]^A_Z X[/itex] and [itex]^A_{Z+1} Y[/itex] are things you can look up in a table of atomic masses. For example, let's take carbon twelve, [itex]^{12}_6 C[/itex]. 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 [itex]^{12}_6 C[/itex] atoms under consideration all have full shells of electrons (in this case 6). If you want to measure the atomic mass of a [itex]^{12}_6 C ^+[/itex] 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, [itex]\Delta m[/itex] 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 [itex]^A_{Z+1} Y[/itex] with the electron subtracted off, and the atomic mass of the beta particle.I'm not sure I understand. [itex]m_e[/itex] is the electron mass.
The byproducts of the beta decay consist of a positive ion and an electron. But the mass of [itex]^A_{Z+1} Y[/itex], 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 [itex]^A_{Z+1} Y[/itex] and subtract off the mass of an electron.

[itex]A_{Z X}[/itex] means [itex]^{12}_6 C[/itex] atom here? [itex]^A_{Z+1} Y[/itex] represent [itex]^{12}_7 C ^+[/itex] ion am i right? to find the mass of [itex]^{12}_6 C[/itex] atom, the mass os 1 electron is subtracted ... why you said that the mass is subtracted to find the mass of [itex]^{12}_5 C ^+[/itex] ion?
 
gxc9800 said:
[itex]A_{Z X}[/itex] means [itex]^{12}_6 C[/itex] atom here?
I was just using carbon twelve as hypothetical example (and in retrospect, not such a good one).

[itex]^A_Z X[/itex] as expressed the problem can be any atom.

[itex]^A_{Z+1} Y[/itex] represent [itex]^{12}_7 C ^+[/itex] ion am i right?

Well, if the parent atom was carbon, then the daughter atom would be a nitrogen ion, [itex]^{12}_7 N^+[/itex]. 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, [itex]^{12}_5 B[/itex], then the daughter ion after a beta decay would be [itex]^{12}_6 C^+[/itex].

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.

[tex]^{14}_6 C \ \ \rightarrow \ \ ^{14}_7 N^+ + e^- + \bar \nu[/tex]

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

to find the mass of [itex]^{12}_6 C[/itex] atom, the mass os 1 electron is subtracted ... why you said that the mass is subtracted to find the mass of [itex]^{12}_5 C ^+[/itex] ion?

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.
 
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 [itex]^{14}_7 N^+[/itex] ion, you could look up the atomic mass of the neutral [itex]^{14}_7 N[/itex] 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 [itex]\Delta m[/itex] because when considering the mass of the beta particle, the same as the mass of the missing electron, they cancel out.