Isotopes of Technetium - Stability of Technetium Isotopes

Stability of Technetium Isotopes

Technetium and promethium are unusual light elements in that they have no stable isotopes. The reason for this is somewhat complicated.

Using the liquid drop model for atomic nuclei, one can derive a semiempirical formula for the binding energy of a nucleus. This formula predicts a "valley of beta stability" along which nuclides do not undergo beta decay. Nuclides that lie "up the walls" of the valley tend to decay by beta decay towards the center (by emitting an electron, emitting a positron, or capturing an electron). For a fixed number of nucleons A, the binding energies lie on one or more parabolas, with the most stable nuclide at the bottom. One can have more than one parabola because isotopes with an even number of protons and an even number of neutrons are more stable than isotopes with an odd number of neutrons and an odd number of protons. A single beta decay then transforms one into the other. When there is only one parabola, there can be only one stable isotope lying on that parabola. When there are two parabolas, that is, when the number of nucleons is even, it can happen (rarely) that there is a stable nucleus with an odd number of neutrons and an odd number of protons (although this happens only in four instances: 2H, 6Li, 10B, and 14N). However, if this happens, there can be no stable isotope with an even number of neutrons and an even number of protons.

For technetium (Z=43), the valley of beta stability is centered at around 98 nucleons. However, for every number of nucleons from 95 to 102, there is already at least one stable nuclide of either molybdenum (Z=42) or ruthenium (Z=44). For the isotopes with odd numbers of nucleons, this immediately rules out a stable isotope of technetium, since there can be only one stable nuclide with a fixed odd number of nucleons. For the isotopes with an even number of nucleons, since technetium has an odd number of protons, any isotope must also have an odd number of neutrons. In such a case, the presence of a stable nuclide having the same number of nucleons and an even number of protons rules out the possibility of a stable nucleus.