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https://www.youtube.com/watch?v=yTkojROg-t8

1) Watch till 0:20

"Then the binding energy is the amount of energy that takes to completely separate them"

But the gravitational force between 2 masses can never be zero.

So the rocket would never be separated from Earth's gravitation which = infinite amount of energy required to separate them.

That means that the gravitational force of Earth has to be infinite.....this assumption should be wrong..

Am I right?

2) Watch from 1:20 to 1:40

If the nucleus has 60 nucleons then why is the radius only 2.5 nucleons?

3) I still don't understand....if energy required to break the bond = the binding energy then how will we get energy when we break the bond? Shouldn't we have to give the energy to break the bond?

Sorry if this is a little vague, if you don't understand what I am talking about then please watch the whole video.

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  • $\begingroup$ No you are right in the first part, in theory the gravitional effect of the Earth goes on and on, it's just so absolutely tiny we can't measure it. $\endgroup$ – user108787 Sep 18 '16 at 13:14
  • $\begingroup$ For number 2 , The stable nucleus has approximately a constant density and therefore the nuclear radius R can be approximated by the following formula, ${\displaystyle R=r_{0}A^{1/3}\,}$ where A = Atomic mass number (the number of protons Z, plus the number of neutrons N) and $r_0$ = 1.25 fm = $1.25 × 10^{−15}$. What radius do you get with this equation with 60 nucleons......? en.wikipedia.org/wiki/Atomic_nucleus $\endgroup$ – user108787 Sep 18 '16 at 13:43
  • $\begingroup$ Please provide enough information to understand your question without having to watch a video on some other site. $\endgroup$ – Ben Crowell Sep 11 at 19:10
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1) One has to be careful with infinities. The concept of binding energy is a quantum mechanical concept, and the gravitational analogue is a bad one, it is classical. In a similar way classically the 1/r potential has an infinity at r=0.

Neither fission nor fusion exist outside a quantum mechanical frame, where there are no infinities because everything is quantized.

2) Again the size of a nucleon is a quantum mechanical concept, each nucleon has a probability density function to exist at (x,y,z,t) so it is not the number of nucleons which will define the size of a nucleus but the way the probability density function aranges these nucleons according to their quantum numbers . The states that the nucleons can exist are quantized and the values depend on the potentials that define the system: the electric charge repulsion of the protons and the attraction of the strong force between baryons ( protons and neutrons). There exist nuclear models for this.

3) If there exist nuclei that could be a subset of a nucleus, if their binding energy per nucleon is larger than of the containing nucleus, for exmaplen the Uranium nucleus, fission can occur (depending again on quantum numbers ) , the U nucleus breaking into the more stable bound subsets. Larger binding energy means energy is given up as it breaks in to the tighter bound fragments

In fusion , two deuterons have a smaller binding energy than an alpha, (look at the binding energy curve) so under appropriate conditions they can fuse into an alpha giving up the extra energy.

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  • $\begingroup$ "...quantum mechanical frame, where there are no infinities because everything is quantized." Seriously? Ever heard of renormalization? $\endgroup$ – flippiefanus Sep 18 '16 at 14:38
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    $\begingroup$ @flippiefanus Sure , it is the mathematics of getting rid of infinities in QM . $\endgroup$ – anna v Sep 18 '16 at 15:02
  • $\begingroup$ @anna v I don't get the 3) part. If the smaller atoms in fission have higher binding evergy/nucleon then wouldn't we have to give in energy? Same goes for fusion $\endgroup$ – MartianCactus Sep 18 '16 at 17:20
  • $\begingroup$ also I don't really get what quantum mechanical frames are. I am a 10 grader :p $\endgroup$ – MartianCactus Sep 18 '16 at 17:23
  • $\begingroup$ Quantum mechanics is the underlying level of all nature, but it is hard to understand it without studying the data and the mathematics that explains it. One is also using E=mc^2 . see here hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html .Maybe you should wait to study the subject at the appropriate class level $\endgroup$ – anna v Sep 18 '16 at 17:56

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