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

In summary, Homework statement states that the daughter nucleus is a positive ion, contrary to the physics fact that says that A_(z+1) Y is a positive nucleus. The atomic masses of ^A_Z X and ^A_{Z+1} Y are things you can look up in a table, but for a ^{12}_6 C ^+ ion, you must subtract off the mass of an electron.
  • #1
gxc9800
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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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  • #2
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 )?
 
  • #3
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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  • #4
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?
 
  • #5
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.
 
  • #6
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.
 

What is beta decay?

Beta decay is a type of nuclear decay process where an unstable nucleus emits a beta particle (either an electron or a positron) and transforms into a more stable nucleus.

What is mass loss in beta decay?

Mass loss in beta decay refers to the decrease in mass of the nucleus as a result of the emission of a beta particle. This decrease in mass is accompanied by the release of energy.

What is the difference between A(z+1)Y and physics fact in relation to beta decay?

A(z+1)Y refers to the atomic mass of the daughter nucleus after beta decay, while physics fact refers to the conservation of energy and momentum in the decay process. The difference between the two is that A(z+1)Y is a measurable quantity, while physics fact is a theoretical concept used to explain the observed behavior of beta decay.

How does beta decay affect the stability of a nucleus?

Beta decay can either increase or decrease the stability of a nucleus depending on the specific type of beta decay (beta-minus or beta-plus) and the resulting daughter nucleus. In general, beta decay helps to stabilize an unstable nucleus by bringing it closer to the line of stability on the chart of nuclides.

What are some practical applications of beta decay?

Beta decay is commonly used in medical imaging and cancer treatment, as well as in nuclear power production. It is also an important tool for studying the properties of nuclei and understanding the processes that occur within stars and during the formation of the universe.

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