Stability of Nuclei

  • #1

Summary:

Apparently binding energies for nuclei are not too much high to be unavailable, but still nuclei don't break during chemical reactions.

Main Question or Discussion Point

Hi
I need help from PF scholars to figure out one difficulty in understanding stability of nucleus. Nuclei remain unaffected during chemical reactions taking place even at very high temperatures and pressures. But their binding energy figures are not that high. Such chemical reactions have heat energies much greater than required binding energies to split the nuclei. For example helium has 28.2 MeV binding energy which is too small. Even sunlight can provide this energy.I am confused why helium or similarly other nuclei don't split at thermal energies? Please help.
 

Answers and Replies

  • #2
Orodruin
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Summary:: Apparently binding energies for nuclei are not too much high to be unavailable, but still nuclei don't break during chemical reactions.

For example helium has 28.2 MeV binding energy which is too small. Even sunlight can provide this energy.
This is simply incorrect. Sunlight corresponds essentially to blackbody radiation with temperature around 6000 K. The typical energy of sunlight photons is therefore in the range of 100s om meV (not MeV, SI prefixes are case sensitive!) - about a factor of ##10^8## too small.

Even the Sun’s core temperature is too low (it is in the keV range).

Edit: ##10^8##, not ##10^5##. Early morning ...
 
  • #3
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I use the rule of thumb that the boundary between the visible and ultraviolet is roughly 3 eV.
 
  • #4
jtbell
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For example helium has 28.2 MeV binding energy which is too small. Even sunlight can provide this energy.
Really? Visible light has wavelengths in the range of about 400 nm to 700 nm. What is the energy of a photon with wavelength e.g. 500 nm?
 
  • #5
Really? Visible light has wavelengths in the range of about 400 nm to 700 nm. What is the energy of a photon with wavelength e.g. 500 nm?
Thanks for the reply. I correct myself. Sure sunlight falls short of binding energy of nucleus. But my question still stands. The thermal energy we provide during chemical reactions aren't sufficient enough to provide this required energy? Please reply with a little bit detail. Regards
 
  • #6
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Thanks for the reply. I correct myself. Sure sunlight falls short of binding energy of nucleus. But my question still stands. The thermal energy we provide during chemical reactions aren't sufficient enough to provide this required energy? Please reply with a little bit detail. Regards
Chemical reactions involve the breaking and forming of bonds. This is where the energy comes from. If you scroll down to the table of bond energies in the hyper link at the end of this, you will see that the bond energies are on the order of electron volts. Now, you can have a huge number of bonds break, and inside your substance you have more than enough energy to cause a nuclear reaction, but this energy is spread out across all the particles. It is averaged out. If you look at a Maxwell-Boltzmann distribution of velocities or kinetic energies at the temperatures achieved in typical chemical reactions you will see that the particles don't get to the energies needed.

http://lamp.tu-graz.ac.at/~hadley/ss1/crystalbinding/bonds/bonds.php
 
  • #7
Thanks Sir. It is very convincing argument. I am pleased and thankful for your time.
High regards.
 
  • #8
I am thankful that the replies have helped me clarify many of my misconceptions. I have one more doubt in mind in support of my statement made earlier regarding breaking of nucleus. In Rutherford nuclear reaction where nitrogen is hit by alpha particles and oxygen and a proton are produced, there is about 1.13 MeV energy required to be possessed by the alpha particles to make it happen. If we mechanically hammer some nitrogen atom, I suppose energy will be much more than 1.13 MeV. Should the nucleus not break by this collision?
1576696887438.png
 
  • #9
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You can do a calculation yourself to see. If you hammer a material all the atoms have the same speed in one direction (ignoring thermal motion). You can calculate the speed you would need for an atom to have that much kinetic energy.
 
  • #10
You can do a calculation yourself to see. If you hammer a material all the atoms have the same speed in one direction (ignoring thermal motion). You can calculate the speed you would need for an atom to have that much kinetic energy.
Thank you Dr_Nate for the kind reply. All I am understanding from your replies is there is some special mechanism by which a nucleus absorb energies. I humbly submit I lack that knowledge and request to kindly share something about it. Is this absorption similar to an electron inside an orbit which absorbs discrete energies only? Please spare a few of your valuable moments. Regards
 
  • #11
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Thank you Dr_Nate for the kind reply. All I am understanding from your replies is there is some special mechanism by which a nucleus absorb energies. I humbly submit I lack that knowledge and request to kindly share something about it. Is this absorption similar to an electron inside an orbit which absorbs discrete energies only? Please spare a few of your valuable moments. Regards
You may be thinking of metastable nuclei that emit gamma rays. Here are two websites. I suggest you go to their homepages and study all the material. You’ll be in a much better position to ask future questions.

Nuclear-power.net
teachnuclear.ca
 
  • #12
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I am thankful that the replies have helped me clarify many of my misconceptions. I have one more doubt in mind in support of my statement made earlier regarding breaking of nucleus. In Rutherford nuclear reaction where nitrogen is hit by alpha particles and oxygen and a proton are produced, there is about 1.13 MeV energy required to be possessed by the alpha particles to make it happen. If we mechanically hammer some nitrogen atom, I suppose energy will be much more than 1.13 MeV. Should the nucleus not break by this collision?
Swing a pickaxe at an anvil. The pickaxe has a plenty of energy to break the iron nucleus that forms the tip of the pickaxe.
The problem is that this energy is divided between several times of 1024 iron nuclei in the axehead. It is not enough to break all nuclei in the axehead. It is even far from enough to break all chemical bonds in the axehead.

There IS a mechanism that will concentrate the widely dispersed kinetic energy of the axehead into the tip of the pickaxe and break the chemical bonds in a small part of the pickaxe. Depending on the detailed properties of the chemical bonds in the tip of the pickaxe either the bonds break and the tip of the pickaxe undergoes plastic flow to go blunt while the eye of the pickaxe is unaltered or else the bonds break and the tip of the pickaxe shatters brittly into splinters, while the eye of the pickaxe is also unaltered but the bonds inside the splinters are also unaltered (only the bonds between splinters break).

But the point is, while there is a mechanism to concentrate the energy to break a small fraction of chemical bonds, there is not a mechanism to concentrate the energy further and break any, even a smaller fraction of nuclei.
 

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