September 25, 2020, 08:47:34 AM
Forum Rules: Read This Before Posting

### Topic: Assumptions made in the Semi-Emperical Mass Formula  (Read 5940 times)

0 Members and 1 Guest are viewing this topic.

#### Mitch ##### Assumptions made in the Semi-Emperical Mass Formula
« on: October 03, 2005, 10:05:39 PM »
What assumption(s) were made when deciding the volume term in the semi-emperical mass formula to be equal to avA.
« Last Edit: October 03, 2005, 10:06:21 PM by Mitch »
Most Common Suggestions I Make on the Forums.
1. Start by writing a balanced chemical equation.
2. Don't confuse thermodynamic stability with chemical reactivity.
3. Forum Supports LaTex

#### Mitch ##### Re:Assumptions made in the Semi-Emperical Mass Formula
« Reply #1 on: October 03, 2005, 10:07:24 PM »
One, is that the radius of the nuclei has a sharp cut-off. Can you think of anything else?
Most Common Suggestions I Make on the Forums.
1. Start by writing a balanced chemical equation.
2. Don't confuse thermodynamic stability with chemical reactivity.
3. Forum Supports LaTex

#### Grejak ##### Re:Assumptions made in the Semi-Emperical Mass Formula
« Reply #2 on: October 03, 2005, 10:27:43 PM »
Two, the nucleus is spherical.

#### alkemist

• Guest ##### Re:Assumptions made in the Semi-Emperical Mass Formula
« Reply #3 on: October 08, 2005, 02:33:29 PM »
Well the major assumption here is that the binding energy is linearly dependent on the number of nucleons, which is what you would expect.  It's more of a guess than an assumption.  Basically, it's assumed that no other factors are important in determining the binding energy of a nucleus other then it's nucleon number, which of course it was proven not to be true, and corrected by other terms that correct for surface energy, the coulombic force,  and symmetry.