Semi-empirical mass formula and fission term

Ryomega
Messages
67
Reaction score
0

Homework Statement



Which term in semi-empirical mass formula gives rise to the energy release in fission? What is the physical origin of this term? All β-stable nuclides heavier than 92Mo (Molybdenum) have a positive Q-value for fission; explain what prevents them decaying in this way.

Homework Equations



Semi-empirical mass formula:

Energy (Binding) = Volume term - Surface term - Coulomb repulsion term - a/symmetry term - pairing term

The Attempt at a Solution



I know that Z2/A ≥ 49 gives spontaneous fission. So that makes the symmetry term responsible for fission. The origin is Pauli exclusion principle where no same states can be occupied.

The rest of the question is complete gibberish to me. I'm not even sure WHCIH way is decaying for what WHAT way is decaying what?

Help.:cry:
 
Physics news on Phys.org
^bump
 
The Q value is the energy released during the reaction. A positive Q-value implies that the reaction may be possible.
However, if you look up the values of Z and A for 92Mo and put them into your formula: \frac{Z^2}{A}\geq49 you will find out why spontaneous fission cannot occur.
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top