Can the 8 e- term be neglected in calculating the energy released in fission?

[Amith2006]

Homework Statement

Consider the following problem:
1) Calculate the energy released in the fission reaction,
Uranium(235,92) + neutron--->[Uranium(235,92)]*--->Neodymium(143,60) + Zirconium(90,40)+ 3 neutrons + 8 electrons +8 antineutrino

Homework Equations

The Attempt at a Solution

Solution:
If atomic rest masses are used in calculating the Q value, the term 8 e- may be dropped.Let M(U),M(Nd),M(Zr) and M(n) be the atomic masses of Uranium, Neodymium,Zirconium and neutron respectively.
Hence,
Q=[M(U) - M(Nd) - M(Zr) - {3-1}M(n)]c^2
Q= 197.6 MeV
This is the solution given in Schaums book on Modern Physics.
Just because atomic rest masses are used, how can we neglect the 8 e- term?

 

[Astronuc]

In the fission process, the electrons are not annihilated, so the rest mass of electrons (92) never changes.

Uranium(235,92) + neutron--->[Uranium(235,92)]*--->Neodymium(143,60) + Zirconium(90,40)+ 3 neutrons + 8 electrons +8 antineutrino

does not have the correct balance.

Fission does not produce protons - it simply transforms one nucleus into two nuclei.

In the fission process, the sum of Z's of the new nuclei (Z1 and Z2) must equal Z(U) = 92, so:

If Z1 = 40 (as in Zr), the other nuclei must have Z2 = 52 (Te).

If Z2 = 60 (as in Nd), then the other nuclei must have Z1 = 32 (Ge).

The number of electrons = Z (which for U = 92).

What does change is the atomic binding energy of the electrons, but that is on the order of eV for valence electrons and increases toward the keV range for inner electrons of more massive nuclei.
See - http://xray.uu.se/hypertext/XREmission.html

 

[Amith2006]

In the fission process, the electrons are not annihilated, so the rest mass of electrons (92) never changes.
Fission does not produce protons - it simply transforms one nucleus into two nuclei.

In the fission process, the sum of Z's of the new nuclei (Z1 and Z2) must equal Z(U) = 92, so:

If Z1 = 40 (as in Zr), the other nuclei must have Z2 = 52 (Te).

If Z2 = 60 (as in Nd), then the other nuclei must have Z1 = 32 (Ge).

The number of electrons = Z (which for U = 92).

What does change is the atomic binding energy of the electrons, but that is on the order of eV for valence electrons and increases toward the keV range for inner electrons of more massive nuclei.

I didn't get your point. Could u please explain it in detail?

 

[stunner5000pt]

I didn't get your point. Could u please explain it in detail?

i think he's saying taht since the binding energy and rest energy of electrons is on the scale of eV and keV we need not care since the answer is on the scale of hundreds of MeV

 

[Amith2006]

i think he's saying taht since the binding energy and rest energy of electrons is on the scale of eV and keV we need not care since the answer is on the scale of hundreds of MeV

But the energy associated with 8 electrons is about 4 MeV and that doesn't seem to negligeble.

 

[Dick]

If you look at charge balance and assume the initial U was neutral, then you will see that the final Nd and Zi can't be neutral. They must be ionized with a net charge of +8. Hence if you use the neutral atomic masses for them, it already accounts for the 8 extra electrons.

 

[Amith2006]

Thats cool!Thanx guys.