4. Binding Energy, the Semi-Empirical Liquid Drop Nuclear Model, and Mass Parabolas
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- Опубликовано: 7 сен 2024
- MIT 22.01 Introduction to Nuclear Engineering and Ionizing Radiation, Fall 2016
Instructor: Michael Short
View the complete course: ocw.mit.edu/22...
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We formally define the binding energy of a nucleus and check our definition with examples from the KAERI Table of Nuclides. We imagine that a nucleus is akin to a droplet of liquid, and construct a semi-empirical mass formula to predict its stability given any number of protons and neutrons. We then construct mass parabolas to explore which nucleus is most stable given a certain number of protons or neutrons. This helps us understand mathematically why certain isotopes undergo which types of radioactive decay, and why certain isotopes are stable.
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At 43:13 you can see this epic, wise, passionate teacher is also a human, and one with politeness and a sense of humor. These lectures are absolutely awesome!
In the differentiation to find the most stable z, there was a mistake. The a_c/(A^(1/3)) term should be negative.
Yes, agree
He had it correct for about 2 secs.
Your both incorrect.
The value was reversed.
A common error that is brought in on the string end
at 9:47 2.79-0.48=2.31MeV
agreed
At 50:10 and a few more times, he misspoke; he meant Ruthenium, not Rubidium.
Does OCW have 22.02? I'm loving this series
Now I know the very low-tech trick of how to easily make a dotted line on a chalkboard.
That will probably be a lost art by 2100.
Liquid Drop Model = hyperfluid vertices in vortices manifestation in/of QM-TIME Superspin, the Periodic Table - Standard Model spectrum of e-Pi-i logarithmic resonance Interference Modulation.
In Perspective Principle Imagery, Superposition-point inside-outside holographic E=mC² mass-energy-momentum positioning -> line-of-sight superposition density-intensity real-numberness pixels for projection-drawing phase-locked coherence-cohesion sync-duration "Droplets" or time-timing sync-duration recirculation resonance node-antinode materialisation.
Atomistic arrangements of Quantum-fields pulse-evolution condensation Actuality are the Phys-Chem cause-effect of time-timing Time Duration Timing Conception. Liquid Drop-> Holographic Universe or Hydrogen Atom bonding arrangements => Eternity-now Quantization Interval Equivalence Principle applies to the Big Picture.
Couldn't have put it better myself.
very cool professor
@MitOpenCourseWare That chalk trick is a Walter Lewin staple. Haha awesome to see him use it!
Yeah I thought the same thing. so cool that he learned it.
25:10- As a liquid droplet, it must comprise molecules as in simplest water droplets which are physically different from protons and neutrons as stated here ? Can't imagine what a proton/neutron only droplet looks like.
The equation on the slide at 06:20 : i + I -> f + F + Q cannot be true since Q = Tf + TF - Ti - TI. I really think it should be i + I -> f + F - Q if we want to keep the same letter notation and Q > 0 as exothermic reaction.
Q can be positive or negative depending on the reaction. So plus is correct.
No, he’s right. Either it should be -Q, or exo and endo are the other way around. Example say in = out + Q, so Q should be in - out, but in his cals he says Q = out - in (Li + He - B, instead of B - Li - He).
15:07 Thorium!
Thanks, thorium guy.
LFTR are the way forward!
Protactinium, Uranium is 92 and Thorium is 90. Come on professor put the spliff down
I got excited about it too! And the professor seemed unsure about its name, even though Th131 is where all the U235 went (there used to be naturally occurring light water nuclear reactors in Gabon a couple billion years ago, as all natural U was "enriched" due to U235's faster half-life), and the reason Radium builds up in basements. Thorium 232 is the "good stuff" you're excited about though!
How to do the chalk trick?
Lewin Walter has explained it in a video
Lewin Walter explains how in this video here: ruclips.net/video/GhawwXcQsUs/видео.html
21:11
Why was he rounding at first? Hm?
Is it just me? Or should it be 2.31MeV and not 2.13MeV? @9:40ish - 10:00ish or am I missing something?
Yes, the teacher corrects it later in the video.
Kindly, let me know how to get lecture notes for this course.
The course materials are available on MIT OpenCourseWare at: ocw.mit.edu/22-01F16. Best wishes on your studies!
Actually, I want to get the power point slides that are used in the class.
Kindly, share those slides.
Thanks
What are those slides he was opening called "Nuclide table".. Actually I mean where to find it?.. Can anybody help?
Not sure if you're looking directly for the PowerPoint slides (if so, I can't find them), but I did find the online nuclide table he uses to gather the information you see in the slides: atom.kaeri.re.kr/old/
For some reason the link on the mit opencourseware website doesn't work...
@@ashleyrshipp What a great resource, thank you! Now I just have to figure out what all the decay modes mean, e.g. SF, IS, IT, as well as J sub Pi, why there are two E sub ex sometimes, etc. I wish there was an index to this thing so I'd know what terms to google and research to better figure out how to use this.
I found this from another thread:
IT stands for Isomeric Transition. A metastable state emits a photon to decay to a lower energy
List of decay modes: ie.lbl.gov/education/decmode.html
“Make sure to conserve mass, energy, and momentum” hem… but mass is not a conserved quantity.
That’s why the concept of excess mass exist.
It is not real. The nucleus doesn’t have “excess mass”. It’s like centrifugal force. We add it to make our equations neat
why do you consider the kinectic energy of the neutron to be zero if you're bombarding Boron with it?
One one side of the reaction you have the initial kinetic energy of the nucleus. On the other side you have the initial kinetic energy of the nucleus plus or minus whatever they gain or lose in the reaction. Since you have it on both sides of the equation you can ignore it or assume it to be zero.
@@1990Judson but isn't the neutron moving with respect to the Boron atom? I thought the neutron was fired at the (mostly) stationary Boron. Maybe I misunderstand how this method works though.
@@jonludwig8233 my thoughts exactly
@@jonludwig8233 This process involves capture of thermal neutrons, i.e. neutrons with energies below 1 eV. This is negligible in comparison with the other energies which are several orders of magnitudes higher (MeV range), so the kinetic energy of the incident neutron can be assumed to be zero.
@@1990Judsonis actually correct.
Another way to think about this is to consider the frame of reference to be the COM of Boron and Neutron before the reaction. Thus, boron will have a negative momentum and neutron will have equal and opposite momentum with net result being zero momentum for the system.
Side note: in any physics problem whether classical, general relative, or quantum it is best to choose the zero momentum reference frame (google it if you don’t know)
Sir,but Coulombic Force is inversely proportional to square of distance!
This is dealing with energy, not force.
One way you can think is that: you have to multiply force with distance to get work which is analogous to energy. That way you can get a value which is inversely proportional to distance.
Or you can think of it as Coulombic voltage. Voltage is interpreted as work done or potential energy gained/lost. Coulombic voltage is inversely proportional to distance.
I am sure these students know the answers, they just don't say. If they don't then I am sure I am way more fitting to sit in that class than them, I am afraid. But you know, I can't yet. Maybe someday, InshAllah.
For some reasons , most students will not answer even though they know the answer, unless encouraged to do so. Some are afraid they will give a wrong answer - even if they perfectly know the right one. I assume someone admitted to MIT is an intelligent and knowledgeable person, so the Dunning - Kruger effect may play also here. Oftentimes the best students have lower self confidence than the average ones. Besides, when the students are asked if there are any questions, they correctly point out the mistake the professor made - not taking into account the gamma energy when calculating T(He) + T(Li) - so you would be correct in not underestimating these guys.
Seems they do know most of the answers. When asked they seem to come up with something intelligent to say.. not sure what you're on about tbh
Way to blow your own trumpet Zippy :D
мать моя женщина...а можно на русском?