What if I show the diagrams of each term and then combine them together?
E.g. for the shape of energy due to strong nuclear force I can say this force has short range characteristics and thus for lightest nuclei the energy due to the strong force grows rapidly. But this trend approaches a limit...
The reverse; an unclear explanation that's correct is not preferable to me!
I'm not.
If you see my first posts, you'll understand that I accepted my mistakes, improved my understanding, and edited my article.
If you consider the post#20, you'll know that I worked hard to improve my...
Just think a 'beginner' with "admitted level of knowledge" wants to read this from Wikipedia.
At first it says ##Q=Ze##; so ##E## 'equals' ##(Ze)^2/(r_0 A^{1/3})##. But because ##Z^2≈Z(Z-1)##, ##E## is 'almost equal to' ##3e^2{Z(Z-1)}/20πε_0r_0 A^{1/3}##.
Doesn't it really mean that ##Z(Z-1)##...
By this statement, I just meant that my understanding of this force is in contrast with what has been said, and thus it is getting 'more and more confusing to me'. I did not mean that "their contributions" are getting worse.
I'm not a native or a good English speaker. I sometimes can't express...
Yes, I know, but I think the explanation of most online sources isn't correct;
For example I've seen Wikipedia, washington.edu, and Lecture 19.
In washington and lecture, it's written that the dependence is on ##Z^2##. And also the way ##Z(Z-1)## is explained in wiki and lecture, can be by...
But I don't agree. As I said we can consider the Coulomb energy of hydrogen as zero, and then use ##Z^2## for other nuclides. As written in Wikipedia & int.washington.edu , a more reliable version is ##Z^2##, but since ##Z^2≈Z(Z-1)## and it goes to zero for hydrogen, we use ##Z(Z-1)##.
I've just decided to remove that formula; because the Liquid Drop Model & SEMF provide a poor (and even wrong) fit for lightest nuclei. And also as you said, if I mention this formula in an introduction article, I should give a motivation and thus a lot of text would be added.
SEMF: Coulomb...
The article I want to publish in PF 'insights blog'. I'm told to share my basic understanding first and ask if they're correct. The post #10 is a small part of that article.
Yes, answering questions is useful, but a thorough understanding of 'fundamental' definitions is a better choice. (For example if you knew what exactly a radioactive decay is, or how to calculate released/utilized energy in nuclear reactions, you would understand this part better.)
Well, the...
I'm glad to hear it!
This is exactly the question written in my article. I know you won't wait until my article gets published, so I send you the related part:
Thus, electric forces do not hold nuclei together, because they act in the opposite direction.
What exactly confines these positively...
I removed that confusing sentence and added some information about Coulomb term instead. Then I designed a more precise diagram by some calculations. @mfb , would you please let me know if there's something wrong (or confusing, unclear, ...)?
In a bound state like atom nucleus (except ##^{1}H##), protons carry a positive charge so repel each other by the Coulomb force. This type of electromagnetic force, (also referred to as electrostatic repulsion) is an inverse-square force (has infinite range characteristics), so a proton added to...
Well, that's my prerogative! Suggesting not to mention something is a suggestion itself!
(I know, I'm a bit stubborn; but the whole my article is based on the definition of binding energy, so I'd rather use that approximation as you said. Thanks for your suggestion -or anything else you call...
Omg, it's getting worse and worse. if it IS the interaction between the quarks of a single nucleon, the fact that "the binding energy of a nucleus is is the energy due to the strong nuclear force, minus the disruptive energy of Coulomb force" is meaningless.
Or if that fact is true, "the strong...
Of course, and I've already said it's an introduction topic.
How about:
Yes, the diagram doesn't indicate a linear relation, and that's exactly what I want; it is due the reason that has already been said.
Ummm, I tried, but couldn't find any formula for the Coulomb force between the protons...
As I said, diagrams don't correspond precise information. I haven't quantified the electrostatic repulsion; I just want to express that because of its infinite range characteristics and small size of nuclei, the disruptive energy of the Coulomb force increases as mass number grows.
I don't mean...
And as I've already included, this article is specially designed for those don't have much information about particle physics, so they do not know what exactly a quark is.
Yes, but more precisely it is the force between the quarks of TWO nucleons.
(The interaction between the quarks of a single nucleon isn't called the strong nuclear force.)
I agree with you, but that's only a small part of my article. (The whole is more than 2000 words!)
As you know, it takes much more time to explain radioactive decay, stability and half life, which needs to be written in another long article! (I may write it after this :smile: )
At first, I want...
Thanks for your reply!
This is almost the last part of my article; before which I've already analyzed BE/A curve and explained exothermic and endothermic processes and how to calculate released/utilizes energy by the difference in final and initial total binding energies (##B_2-B_1##). Then I...
In fusion reactions, lighter nuclei are combined into heavier ones (atomic mass increases); so the direction is from left to right (shown on the diagram). Therefore, in the fusion of 'light elements' the total final binding energy is higher (the related arrow points upward); so ##B_2-B_1>0## and...
Exothermic process is a [nuclear] reaction in which energy is released, but for endothermic processes energy is needed. What does this released or needed energy come from?
Let's say we have the fusion shown in this picture:
##m_h, m_t, m_α, m_p, m_n## - masses of helium-3, tritium, helium-4...
The best-known classes of nuclear transmutations are fission and fusion:
Nuclear Fusion
Under normal condition, nuclei do not stick together; because they repel each other at large distances (due to the electrostatic repulsion 'barrier') and thus the strong nuclear force cannot act. But if these...