# Homework Help: Energy released in fission

1. May 28, 2007

### Amith2006

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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?

2. May 28, 2007

### Astronuc

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

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 [Broken]

Last edited by a moderator: May 2, 2017
3. May 29, 2007

### Amith2006

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

4. May 29, 2007

### stunner5000pt

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

5. May 31, 2007

### Amith2006

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

6. Jun 1, 2007

### 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.

7. Jun 1, 2007

### Amith2006

Thats cool!Thanx guys.