I have a question on the "decay" of Vanadium-48. The reason it's in inverted commas is because I'm not sure whether decay is the right word.

Basically what I'm trying to work out is whether it's likely/possible that I can, via a few steps, get from V-48 to Sc-46.

Is it possible that if I start with a large amount of V-48, this can over time "decay" into V-45? Can Isotopes of elements just lose neutrons and change into other isotopes?

After that I'm hoping that I can follow the natural decay of V-45 into Ti-45 via electron capture, and again via electron capture this goes to Sc-45, which perhaps via Neutron capture could become Sc-46 with gamma emission?

Is my thinking on this completely wrong? Does my question make sense? Appreciate any help or advice, I'm sure it's relatively obvious that I'm no expert.

Many thanks.


1 Answer 1


According to wikipedia, Vanadium-48 decays via $\beta^+$ (positron emission) to Titanium-48, which is a stable isotope.

The emission of neutrons for Vanadium-48 isn't allowed becuase it doesn't conserve energy: Vanadium-48 has a mass of 47.9522537 u, and Vanadium-45 plus 3 neutrons have a total mass of 44.965776 u +3·1.00866491600 u = 47.991770 u. For a nuclear reaction to be allowed, the mass of the daughter nuclei must be lower than the mass of the father nucleus.

Note that there are some isotopes where neutron emission is energetically allowed, such as Helium-5 or Beryllium-13. But that's not the case for Vanadium-48.


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