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Q & A: Stability of He-3 and Tritium?

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Most recent answer: 08/26/2011
Q:
Why is helium-3 stable whereas tritium decays with a short half-life?
- Michael
Milwaukee, WI
A:
  It all has to to do with the conservation of energy and the quantum mechanical dictum 'If something is allowed then it will eventually happen'. 
 If you add up the total energy of a He-3 nucleus, including the rest mass of the constituent protons and neutron as well as the nuclear binding energy, then you will find it is less than that of any possible decay products, for example a deuterium nucleus plus proton.  Energy cannot be conserved therefore this decay is forbidden.  However, the energy of a tritium nucleus is more than that of a He-3 nucleus plus an electron and neutrino.  So it will cheerfully decay into He-3 via a virtual beta decay of neutron to a proton, electron and neutrino. The resulting kinetic energy of the decay products is 18.6 KeV.

LeeH

(published on 12/17/2010)

Follow-Up #1: Where's the missing energy in Tritium decay?

Q:
While it is true that the mass of He-3 is less than that of tritium, the mass of He-3 plus the mass of an electron (which is generated from the beta decay of tritium) is actually greater than the mass of tritium. The argument that you give doesn't explain this. The mass values that I'm using are: He-3: 3.0160293u Tritium: 3.0160492u electron: 0.0005486u I am somewhat baffled by the fact that the energy difference between He-3 and Tritium is almost exactly 18.6 keV, the energy of the beta particle. It seems to close to be a coincidence... I'm interested to hear your thoughts on this.
- Mike (age 29)
Nashville, TN, USA
A:
Actually, there is no missing energy.   The atomic mass of He3 already includes the extra electron.  The energetics of the decay is driven by the difference in binding energy of the two nuclei.  The two protons in He3 tend to repel each other due to their like-charges thus reducing its binding energy relative to that of the two neutrons in tritium, H3.

LeeH

(published on 08/26/2011)

Follow-up on this answer.